Chapter 10 treats the Discrete Fourier Transform as a matrix. The reason that matrix is worth studying through its eigenvalues, rather than just multiplying it out numerically, is exactly the machinery this section reviews. Nothing here is new. Eigenvalues, eigenvectors, and diagonalizing a symmetric matrix into an orthogonal eigenbasis are all Linear Algebra I results. What's new is watching, ahead of time, how directly that orthogonal-eigenbasis guarantee reaches into a course about frequencies.
An eigenvector is a direction a matrix treats like a one-way street with no curves. Travel along it, and A only changes your speed, by a factor of \lambda, never your heading. Most directions aren't like this at all. A bends them onto some new heading entirely. A symmetric matrix is unusually generous. It always hands you a full set of these straight streets. Better still, those streets are guaranteed to run perpendicular to each other, the fact this course keeps needing. Not merely straight, but straight and mutually at right angles, the same orthogonality the last two sections built for functions instead of vectors.
The trap with a repeated eigenvalue is treating the repeat itself as bad news. A repeated eigenvalue is a warning to go check something, not a verdict. Solve (A-\lambda I)v=\mathbf 0 for that eigenvalue and see how many independent directions actually solve it. If that eigenspace has full size, the repeat causes no trouble whatsoever. The matrix hands over exactly as many independent eigenvectors from that one eigenvalue as it needed to. Only an eigenspace that comes up short actually breaks diagonalizability.
A nonzero vector v is an eigenvector of a square matrix A if Av = \lambda v for some scalar \lambda, its matching eigenvalue. It's a direction A only stretches or shrinks, never rotates off its own line. Eigenvalues are found from the characteristic equation \det(A-\lambda I) = 0. Each root \lambda then has its eigenvectors recovered by solving (A-\lambda I)v = \mathbf 0.
A is diagonalizable, meaning A = PDP^{-1} for invertible P and diagonal D, exactly when \mathbb R^n has a full basis made of A's eigenvectors (the columns of P), with D's diagonal holding the matching eigenvalues in the same order. Diagonalization is the process of finding that P and D. A repeated eigenvalue doesn't automatically break this. What matters is whether that eigenvalue's eigenspace (the solution set of (A-\lambda I)v=\mathbf 0) has enough dimension to supply its full share of independent eigenvectors. When A is symmetric (A=A^{\mathsf T}), this always works out. Every real symmetric matrix is diagonalizable, its eigenvalues are guaranteed real, and its eigenvectors from different eigenvalues are automatically orthogonal. That lets D=Q^{\mathsf T}AQ for an orthogonal matrix Q, instead of a merely invertible P.
Theorem
The Diagonalization Theorem
Recalled from Linear Algebra — Vectors, Matrices, and the Geometry of Data · Diagonalization — first taught there.
An n \times n matrix A is diagonalizable — that is, A = PDP^{-1} for some invertible P and diagonal D — if and only if A has n linearly independent eigenvectors. When this holds, the columns of P are those eigenvectors and the matching diagonal entries of D are their eigenvalues.
(\Leftarrow) n independent eigenvectors give A = PDP^{-1}. Suppose v_1, \dots, v_n are linearly independent eigenvectors of A, with Av_i = \lambda_i v_i. Let P be the matrix with these vectors as its columns, and let D = \operatorname{diag}(\lambda_1, \dots, \lambda_n). Multiplying A into P column by column, AP = A\begin{bmatrix} v_1 & \cdots & v_n \end{bmatrix} = \begin{bmatrix} Av_1 & \cdots & Av_n \end{bmatrix} = \begin{bmatrix} \lambda_1 v_1 & \cdots & \lambda_n v_n \end{bmatrix}. Multiplying P by D on the right scales each column of P by the matching diagonal entry of D, producing the very same matrix: PD = \begin{bmatrix} v_1 & \cdots & v_n \end{bmatrix} \operatorname{diag}(\lambda_1,\dots,\lambda_n) = \begin{bmatrix} \lambda_1 v_1 & \cdots & \lambda_n v_n \end{bmatrix}. So AP = PD. Because P's columns are n linearly independent vectors in \mathbb{R}^n, P is invertible, and right-multiplying both sides by P^{-1} gives A = PDP^{-1}.
(\Rightarrow) A = PDP^{-1} gives n independent eigenvectors. Suppose A = PDP^{-1} with P invertible and D = \operatorname{diag}(\lambda_1, \dots, \lambda_n). Then AP = PD, and reading that equation column by column exactly as above shows Av_i = \lambda_i v_i, where v_i is the i-th column of P. Each v_i is nonzero — a zero column would make P singular — so v_1,\dots,v_n are eigenvectors of A. Since P is invertible, its columns are linearly independent, which produces the needed n independent eigenvectors.
Sanity check: take A = \begin{bmatrix} 4 & 1 \\ 0 & 3 \end{bmatrix}, upper triangular, so its eigenvalues sit on the diagonal: \lambda_1 = 4, \lambda_2 = 3. Solving (A - 4I)v = 0 gives eigenvector v_1 = (1,0), and (A-3I)v = 0 gives v_2 = (1,-1). These two are not scalar multiples of each other, so they're independent, and the theorem guarantees A = PDP^{-1} with P = \begin{bmatrix} 1 & 1 \\ 0 & -1 \end{bmatrix} and D = \begin{bmatrix} 4 & 0 \\ 0 & 3 \end{bmatrix} — which a direct multiplication PDP^{-1} confirms reproduces A.
Theorem
Symmetric Matrices Are Orthogonally Diagonalizable
Recalled from Linear Algebra — Vectors, Matrices, and the Geometry of Data · Symmetric matrices and orthogonal diagonalization — first taught there.
Let A be an n \times n real matrix. Then A is symmetric — A = A^{\mathsf T} — if and only if there is an orthogonal matrix Q and a diagonal matrix D with A = QDQ^{\mathsf T}. The columns of Q form an orthonormal basis of \mathbb{R}^n made of eigenvectors of A, and the diagonal entries of D are the matching eigenvalues.
(\Leftarrow) Orthogonal diagonalizability forces symmetry. Suppose A = QDQ^{\mathsf T} with Q orthogonal and D diagonal. Diagonal matrices equal their own transpose, and transposing a product reverses the order of its factors, so A^{\mathsf T} = (QDQ^{\mathsf T})^{\mathsf T} = (Q^{\mathsf T})^{\mathsf T}D^{\mathsf T}Q^{\mathsf T} = QDQ^{\mathsf T} = A. So A is symmetric. The rest of the proof handles the harder direction.
Symmetric matrices have only real eigenvalues. Let \lambda be an eigenvalue of A, allowing for the moment that it might be complex, with eigenvector v (possibly with complex entries), so Av = \lambda v. Let \bar v denote the entrywise complex conjugate of v, and consider the scalar \bar v^{\mathsf T} A v. Computing it one way, \bar v^{\mathsf T} A v = \bar v^{\mathsf T}(\lambda v) = \lambda (\bar v^{\mathsf T} v) = \lambda \lVert v \rVert^2, since \bar v^{\mathsf T} v = \sum_i \overline{v_i} v_i = \sum_i |v_i|^2 = \lVert v \rVert^2 is real and positive (v \neq 0, being an eigenvector). Computing the same scalar the other way, using A^{\mathsf T} = A and that \bar v^{\mathsf T} A v is a 1\times 1 matrix and so equal to its own transpose, \bar v^{\mathsf T} A v = \left(\bar v^{\mathsf T} A v\right)^{\mathsf T} = v^{\mathsf T} A^{\mathsf T} \bar v = v^{\mathsf T} A \bar v. Taking the complex conjugate of Av = \lambda v gives A\bar v = \bar\lambda \bar v (since A has real entries, conjugation passes through it untouched), so v^{\mathsf T} A \bar v = v^{\mathsf T}(\bar\lambda \bar v) = \bar\lambda (v^{\mathsf T}\bar v) = \bar\lambda \lVert v\rVert^2, using v^{\mathsf T}\bar v = \overline{\bar v^{\mathsf T} v} = \lVert v\rVert^2 again. The two computations of \bar v^{\mathsf T} A v must agree: \lambda \lVert v \rVert^2 = \bar\lambda \lVert v \rVert^2. Since \lVert v \rVert^2 \neq 0, this forces \lambda = \bar\lambda, i.e. \lambda is real. So every eigenvalue of A is real, and consequently every eigenvalue has a real eigenvector (a real matrix minus a real multiple of I_n has a real null space).
Eigenvectors for distinct eigenvalues are orthogonal. Let Av_1 = \lambda_1 v_1 and Av_2 = \lambda_2 v_2 with \lambda_1 \neq \lambda_2. Using symmetry, \lambda_1 (v_1 \cdot v_2) = (Av_1) \cdot v_2 = v_1^{\mathsf T}A^{\mathsf T} v_2 = v_1^{\mathsf T}Av_2 = v_1 \cdot (Av_2) = \lambda_2(v_1 \cdot v_2). So (\lambda_1 - \lambda_2)(v_1 \cdot v_2) = 0, and since \lambda_1 \neq \lambda_2, this forces v_1 \cdot v_2 = 0.
Building Q by induction on n. The base case n=1 is trivial: any 1\times 1 matrix is already diagonal, and Q = [1] is orthogonal. For the inductive step, assume every symmetric (n-1)\times(n-1) matrix is orthogonally diagonalizable, and let A be a symmetric n \times n matrix. By the real-eigenvalue fact above, A has a real eigenvalue \lambda_1 with a real, nonzero eigenvector; rescale it to a unit vector q_1, so Aq_1 = \lambda_1 q_1 and \lVert q_1 \rVert = 1. Extend \{q_1\} to an orthonormal basis \{q_1, q_2, \dots, q_n\} of \mathbb{R}^n (extend to any basis and apply Gram–Schmidt, which leaves q_1 fixed since it is already a unit vector orthogonal to nothing else yet chosen), and let P be the orthogonal matrix with these vectors as columns.
The subspace W = \operatorname{span}(q_2,\dots,q_n) = q_1^\perp is invariant under A: for w \in W, (Aw)\cdot q_1 = w \cdot (Aq_1) = w \cdot (\lambda_1 q_1) = \lambda_1(w\cdot q_1) = 0, using symmetry in the first step exactly as above, so Aw is again orthogonal to q_1, i.e. Aw \in W. Consequently P^{\mathsf T}AP has the block form P^{\mathsf T}AP = \begin{bmatrix} \lambda_1 & 0 \\ 0 & A_1 \end{bmatrix} for some symmetric (n-1)\times(n-1) matrix A_1: the first column of P^{\mathsf T}AP is P^{\mathsf T}(Aq_1) = P^{\mathsf T}(\lambda_1 q_1) = \lambda_1 e_1 (since P^{\mathsf T}q_1 = e_1, q_1 being the first column of the orthogonal matrix P), giving the zero block below \lambda_1; and P^{\mathsf T}AP is symmetric because A is, which forces the zero block to the right of \lambda_1 as well. By the inductive hypothesis, A_1 = Q_1D_1Q_1^{\mathsf T} for some orthogonal (n-1)\times(n-1) matrix Q_1 and diagonal D_1. Setting Q = P\begin{bmatrix} 1 & 0 \\ 0 & Q_1 \end{bmatrix}, \qquad D = \begin{bmatrix} \lambda_1 & 0 \\ 0 & D_1 \end{bmatrix}, the block-diagonal factor is orthogonal (its columns are orthonormal, being e_1 together with the orthonormal columns of Q_1), so Q is a product of orthogonal matrices and hence orthogonal itself. Unwinding the block product, Q^{\mathsf T}AQ = \begin{bmatrix} 1 & 0 \\ 0 & Q_1^{\mathsf T} \end{bmatrix}P^{\mathsf T}AP\begin{bmatrix} 1 & 0 \\ 0 & Q_1 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & Q_1^{\mathsf T} \end{bmatrix}\begin{bmatrix} \lambda_1 & 0 \\ 0 & A_1 \end{bmatrix}\begin{bmatrix} 1 & 0 \\ 0 & Q_1 \end{bmatrix} = \begin{bmatrix} \lambda_1 & 0 \\ 0 & Q_1^{\mathsf T}A_1Q_1 \end{bmatrix} = D, so A = QDQ^{\mathsf T} with Q orthogonal and D diagonal, completing the induction.
Sanity check: take the symmetric matrix A = \begin{bmatrix} 2 & 1 \\ 1 & 2 \end{bmatrix}. Its characteristic equation is (2-\lambda)^2 - 1 = 0, so \lambda_1 = 3 and \lambda_2 = 1 — both real, as guaranteed. Solving (A-3I)v=0 gives v_1 = (1,1), and (A-I)v=0 gives v_2 = (1,-1); these two are orthogonal (v_1 \cdot v_2 = 1 - 1 = 0), exactly as the theorem predicts for eigenvectors of distinct eigenvalues. Normalizing gives q_1 = \frac{1}{\sqrt2}(1,1) and q_2 = \frac{1}{\sqrt2}(1,-1), and setting Q = \begin{bmatrix} 1/\sqrt2 & 1/\sqrt2 \\ 1/\sqrt2 & -1/\sqrt2 \end{bmatrix}, D = \begin{bmatrix} 3 & 0 \\ 0 & 1 \end{bmatrix}, a direct multiplication confirms QDQ^{\mathsf T} = A.
Hunting for a symmetric matrix's two perpendicular directions
Drag v around the unit circle. Av snaps onto v's own ray at exactly two spots, and those two spots sit at a right angle to each other.
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Recipe Find a matrix's eigenvalues and determine whether it's diagonalizable Use when: you have a square matrix A and need its eigenvalues, plus a settled yes or no call on whether A=PDP^{-1} for some invertible P Goal: every eigenvalue of A, plus a verdict on whether A has enough independent eigenvectors to diagonalize
Quick recipe
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1
Subtract \lambda from each diagonal entry only, then set the determinant of the result equal to zero.
Example
For A=\begin{pmatrix}5&2\\2&5\end{pmatrix}: A-\lambda I=\begin{pmatrix}5-\lambda&2\\2&5-\lambda\end{pmatrix}, so \det(A-\lambda I)=(5-\lambda)^2-4=0.
Why: an eigenvalue is exactly a \lambda that makes A-\lambda I singular, sending some nonzero v to \mathbf 0.
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2
Expand and solve for \lambda. Every root found this way is an eigenvalue of A.
Example
(5-\lambda)^2-4=0 \Rightarrow (5-\lambda)^2=4 \Rightarrow 5-\lambda=\pm2, giving \lambda=3 or \lambda=7, two distinct roots.
- every root turns out distinct — each eigenvalue automatically gets a full, independent eigenvector; skip to Step 4.
- a root repeats — continue to Step 3 before drawing any conclusion about diagonalizability.
⚡ Shortcut
For any 2\times2 matrix, skip straight to \lambda=\frac{\operatorname{tr}A\pm\sqrt{(\operatorname{tr}A)^2-4\det A}}{2}. Here \operatorname{tr}A=5+5=10 and \det A=25-4=21, so \lambda=\frac{10\pm\sqrt{100-84}}{2}=\frac{10\pm4}{2}=7 or 3, matching the long way exactly.
Why: the roots are the only candidates for eigenvalues; the shortcut only reaches the same equation faster for a 2\times2 matrix.
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3
Find how many independent vectors solve the homogeneous system for that repeated \lambda. If that count matches the eigenvalue's multiplicity, it contributes enough eigenvectors; if it falls short, diagonalization fails outright.
Example
For D=\begin{pmatrix}4&0\\0&4\end{pmatrix} with repeated \lambda=4: (D-4I)v=\mathbf 0 is \mathbf 0v=\mathbf 0, so every vector works, a full 2-dimensional eigenspace. For E=\begin{pmatrix}4&1\\0&4\end{pmatrix}, also repeated \lambda=4: (E-4I)v=\mathbf 0 forces the second coordinate to 0, leaving only a 1-dimensional eigenspace, one short.
- the eigenspace's dimension matches the eigenvalue's multiplicity — this eigenvalue is fine; check whether every other eigenvalue also has enough.
- the eigenspace's dimension falls short of the multiplicity — A is not diagonalizable, and no other eigenvalue can fix that.
Why: P needs a full set of n independent eigenvector columns to be invertible, and a repeated eigenvalue only threatens that when its own eigenspace comes up short.
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4
Declare A diagonalizable once every eigenvalue supplied its full share of independent eigenvectors. If A is also symmetric, those eigenvectors are automatically orthogonal across different eigenvalues.
Example
For A=\begin{pmatrix}5&2\\2&5\end{pmatrix} (symmetric), the eigenvectors are v_1=(1,1) for \lambda=7 and v_2=(1,-1) for \lambda=3: v_1\cdot v_2=(1)(1)+(1)(-1)=0, orthogonal with no extra effort.
- A is symmetric — eigenvectors from different eigenvalues are guaranteed orthogonal, with no extra work.
- A is diagonalizable but not symmetric — independence still holds, but perpendicularity is not guaranteed; check it directly if it's needed.
Why: orthogonality is a bonus that symmetry specifically guarantees, not a property every diagonalizable matrix's eigenvectors share.
Common mistakes
- Assuming a repeated eigenvalue always blocks diagonalization, when what actually matters is the eigenspace's dimension.
- Expecting eigenvectors to be orthogonal from a matrix that isn't symmetric.
- Using the trace-determinant shortcut on a 3\times3 or larger matrix, where it doesn't collapse to one quadratic.
- Stopping at the characteristic equation without actually solving it for every root.
Worked example
Finding eigenvalues from the characteristic equation
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1
Find the eigenvalues of A=\begin{pmatrix}3&1\\1&3\end{pmatrix}, running Steps 1 and 2 of the recipe above on a fresh matrix.
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2\det(A-\lambda I) = \det\begin{pmatrix}3-\lambda & 1\\1&3-\lambda\end{pmatrix} = (3-\lambda)^2 - 1.
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3
Set this to 0: (3-\lambda)^2=1, so 3-\lambda=\pm1, giving \lambda = 2 or \lambda=4.
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4
\boxed{\lambda = 2, 4}. Both real, as symmetry guarantees.
Worked example
A shortcut: eigenvalues from trace and determinant alone
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1
Recompute A=\begin{pmatrix}3&1\\1&3\end{pmatrix}'s eigenvalues without expanding \det(A-\lambda I) by hand, using the Step 2 shortcut from the recipe above.
For any 2\times2 matrix, the characteristic equation always simplifies to \lambda^2 - (\operatorname{tr}A)\lambda + \det A = 0, so the quadratic formula gives eigenvalues directly: \lambda = \frac{\operatorname{tr}A \pm \sqrt{(\operatorname{tr}A)^2-4\det A}}{2}.
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2
\operatorname{tr}A = 3+3=6, \det A = 9-1=8.
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3\lambda = \frac{6\pm\sqrt{36-32}}{2} = \frac{6\pm2}{2} = 4 \text{ or } 2.
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4
\boxed{\lambda=2,4}, matching the full determinant expansion in fewer steps. This shortcut is genuinely only a 2\times2 fact. A 3\times3 or larger characteristic polynomial doesn't collapse to a single quadratic, and the full \det(A-\lambda I)=0 expansion is needed instead.
Assembling A=PDP^{-1} from its own eigenpairs
Reveal each eigenpair. Watch P, D, and the checked product PDP^{-1} build up until it matches A entrywise.
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Worked example
Confirming the eigenvectors are orthogonal
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1
Using the eigenvectors from above, v_1=(1,1) for \lambda=4 and v_2=(1,-1) for \lambda=2, confirm they're orthogonal, the same Step 4 check the recipe above runs on a symmetric matrix.
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2v_1\cdot v_2 = (1)(1)+(1)(-1) = 0.
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3
\boxed{v_1\perp v_2}, exactly as the Symmetric Matrices theorem guarantees for eigenvectors from different eigenvalues of a symmetric matrix, with no extra work needed to arrange it.
Worked example
A repeated eigenvalue that IS diagonalizable, and one that isn't
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1
Compare B=\begin{pmatrix}2&0\\0&2\end{pmatrix} and C=\begin{pmatrix}2&1\\0&2\end{pmatrix}, both with the repeated eigenvalue \lambda=2.
Common trap. Both have the same repeated eigenvalue, which invites assuming both fail, or both succeed, at diagonalization together. Step 3 of the recipe above exists exactly to settle that question case by case.
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2
For B: solve (B-2I)v=\mathbf 0, i.e. \begin{pmatrix}0&0\\0&0\end{pmatrix}v=\mathbf 0. Every vector in \mathbb R^2 works, so the eigenspace is 2-dimensional, a full basis on its own.
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3
For C: solve (C-2I)v=\mathbf 0, i.e. \begin{pmatrix}0&1\\0&0\end{pmatrix}v=\mathbf 0. This forces the second coordinate to be 0, leaving only multiples of (1,0). That's a 1-dimensional eigenspace, one short of the two independent eigenvectors diagonalization needs.
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4
\boxed{B\text{ is diagonalizable (trivially, }B=2I\text{ already); }C\text{ is not}}. The repeated eigenvalue was identical in both. Only the eigenspace's actual dimension decided the outcome. (C isn't symmetric, which is exactly why the guarantee above doesn't cover it.)
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Where this goes next
Diagonalizing the DFT matrix, chapter 10
The Discrete Fourier Transform is, underneath its formula, multiplication by a specific matrix. Chapter 10 shows that a whole family of matrices, the circulant matrices that model any operation built from shifting a signal, are all diagonalized by that same DFT matrix's eigenvectors. "Diagonalizable, with a shared eigenbasis across a whole family of matrices" is the underlying structure. Recognizing it is precisely the Linear Algebra I fact this section just reviewed. Only the matrix is bigger now, built from complex exponentials instead of small integers.
Check your understanding (counts toward your progress)
A 3\times3 matrix has a repeated eigenvalue. Is it automatically non-diagonalizable?
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A repeated eigenvalue only matters if its eigenspace comes up short of the repeat count — a repeat with a full-dimensional eigenspace causes no problem at all.
Why
Diagonalizability depends on whether every eigenvalue's eigenspace dimension matches its multiplicity — a repeat is only a problem when that dimension falls short.
Compare the eigenvalue's algebraic multiplicity (how many times it repeats) to the dimension of its eigenspace.
For a real symmetric matrix, eigenvectors from two DIFFERENT eigenvalues are always ___.
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Why
Symmetric matrices guarantee that eigenvectors belonging to distinct eigenvalues are orthogonal to each other — part of what makes an orthogonal diagonalization A=QDQ^{\mathsf T} possible.
Find the smaller eigenvalue of \begin{pmatrix}5&0\\0&2\end{pmatrix}.
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Why
A diagonal matrix's eigenvalues are just its diagonal entries: 5 and 2.
Correct answer
Also accepted:
Why
Practice: Eigenvectors, eigenvalues, and when diagonalization works (4 problems · doesn’t affect your progress)
Determine the smaller of the two eigenvalues of \begin{pmatrix}6&2\\2&3\end{pmatrix}.
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Why
\operatorname{tr}A=9, \det A=18-4=14, so \lambda=\frac{9\pm\sqrt{81-56}}{2}=\frac{9\pm5}{2}=7 or 2; the smaller one is 2.
Correct answer
Also accepted:
Why
For a 2\times2 matrix, \lambda=\frac{\operatorname{tr}A\pm\sqrt{(\operatorname{tr}A)^2-4\det A}}{2} skips the determinant expansion.
Is v=(1,-1) an eigenvector of A=\begin{pmatrix}2&1\\1&2\end{pmatrix}?
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Recompute Av directly — \lambda=3 is the eigenvalue that belongs to a DIFFERENT eigenvector of this matrix.
Av does come out as a scalar multiple of v here — check the arithmetic again.
Why
Av = (2(1)+1(-1),\ 1(1)+2(-1)) = (1,-1) = 1\cdot v, so v is an eigenvector with eigenvalue 1.
Yes, with eigenvalue 1
A 3\times3 matrix has eigenvalues 2, 2, 5, and the eigenspace for \lambda=2 turns out to be only 1-dimensional. Is the matrix diagonalizable?
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\lambda=2 repeats twice, so diagonalization needs a 2-dimensional eigenspace there — a 1-dimensional one is one short.
Why
\lambda=2 has multiplicity 2 but only a 1-dimensional eigenspace, so there aren't enough independent eigenvectors to build a full eigenbasis — not diagonalizable.
No — the eigenspace for \lambda=2 falls short of its multiplicity
A real symmetric matrix has eigenvectors v_1=(2,1) and v_2=(a,-2) from two different eigenvalues. What must a equal?
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Why
Eigenvectors from different eigenvalues of a symmetric matrix are guaranteed orthogonal, so v_1\cdot v_2=2a-2=0, giving a=1.
Correct answer
Also accepted:
Why