Linear-Algebra-Done-Right · Unit 19 · Video 1 · Interactive Practice

Real Spectral Theorem: Self-Adjoint ⇔ Orthonormal Eigenbasis

IKey Formulas

Formula	Name	What it says
$x^2 + bx + c = \left(x + \tfrac{b}{2}\right)^2 + \left(c - \tfrac{b^2}{4}\right) > 0$	Completing the square	For real $b$ , $c$ with $b^2 < 4c$ and every real $x$ : the quadratic has no real zero
$\langle (T^2 + bT + cI)v, v \rangle \ge \left(\\|Tv\\| - \tfrac{\|b\|\,\\|v\\|}{2}\right)^2 + \left(c - \tfrac{b^2}{4}\right)\\|v\\|^2 > 0$	Invertible quadratic expressions	For $T = T^*$ , $b^2 < 4c$ and $v \neq 0$ : so $T^2 + bT + cI$ is injective, hence invertible
minimal polynomial of $T$ $= (z - \lambda_1)\cdots(z - \lambda_m)$ , $\lambda_k \in \mathbf{R}$	Minimal polynomial of a self-adjoint operator	Every quadratic factor $z^2 + bz + c$ with $b^2 < 4c$ would be invertible at $T$ and could be cancelled; so $T$ has an eigenvalue even when $\mathbf{F} = \mathbf{R}$
$\mathbf{F} = \mathbf{R}$ : $T = T^* \iff \mathcal{M}(T)$ diagonal w.r.t. some orthonormal basis $\iff$ $V$ has an orthonormal basis of eigenvectors of $T$	Real spectral theorem	Diagonalizable alone is weaker: the eigenvectors must also be orthonormal

Key Insight: The minimal polynomial gives an upper-triangular matrix with respect to an orthonormal basis; $T = T^*$ makes that matrix equal its transpose, and only the diagonal survives.

IIWhy $T^2 + bT + cI$ Cannot Send $v \neq 0$ to $0$

With $Q = T^2 + bT + cI$ , $T = T^*$ and $b^2 < 4c$ , completing the square forces $\langle Qv, v \rangle > 0$ for every unit vector $v$ .

Operator b 1 c 1 angle of v 35°

💡 Challenge: find $b$ , $c$ with $b^2 \ge 4c$ that make $Qv = 0$ for every $v$ when $T$ is self-adjoint. Then $z^2 + bz + c$ is the minimal polynomial of $T$ — and its zeros are real, as the theorem demands.

IIISymmetric Matrix, Perpendicular Eigenvectors

A real $2 \times 2$ matrix with $\mathcal{M}^{t} = \mathcal{M}$ always has perpendicular eigenvector lines; without symmetry they tilt, merge or vanish.

Matrix entry (1,1) 1 entry (1,2) 2 entry (2,1) 2 entry (2,2) −2

IV(1) ⇒ (2): Upper-Triangular and Self-Adjoint Leaves Only the Diagonal

With respect to an orthonormal basis $\mathcal{M}(T^*) = \mathcal{M}(T)^{t}$ , so an upper-triangular matrix of a self-adjoint $T$ is diagonal.

Step 1

VQuiz Questions

Problem 1 · An Orthonormal Eigenbasis in the Plane

Given: $T \in \mathcal{L}(\mathbf{R}^2)$ has matrix $A = \begin{pmatrix} 2 & 2 \\ 2 & 5 \end{pmatrix}$ with respect to the standard basis, and $\mathbf{R}^2$ has the Euclidean inner product. Find an orthonormal basis of $\mathbf{R}^2$ consisting of eigenvectors of $T$ .

✅ Correct!

T(2, -1) = (2, -1)

and

T(1, 2) = (6, 12) = 6(1, 2)

; the two vectors are orthogonal and each has length

\sqrt{5}

, so in this basis

\mathcal{M}(T) = \operatorname{diag}(1, 6)

❌ Not quite. The standard basis is orthonormal, but

T(1, 0) = (2, 2)

is not a multiple of

(1, 0)

A

is not diagonal, so these are not eigenvectors. The theorem promises some orthonormal eigenbasis, usually not the one you started with.

❌ Not quite.

T(1, -2) = (2 - 4,\ 2 - 10) = (-2, -8)

is not a multiple of

(1, -2)

, and

(1, 2) \cdot (1, -2) = -3 \neq 0

, so the pair is not even orthogonal. The eigenvalue-

1

eigenvector solves

v_1 + 2v_2 = 0

, giving

(2, -1)

❌ Not quite. These are orthogonal eigenvectors (eigenvalues

1

and

6

), but each has length

\sqrt{5}

, not

1

. An orthonormal basis needs unit vectors: divide each by

\sqrt{5}

Show solution

$A = A^{t}$ and the standard basis is orthonormal, so $T = T^*$ , and the real spectral theorem guarantees an orthonormal eigenbasis.

Eigenvalues. $A^2 = \begin{pmatrix} 8 & 14 \\ 14 & 29 \end{pmatrix} = 7A - 6I$ , and $A$ is not a multiple of $I$ , so the minimal polynomial is $z^2 - 7z + 6 = (z - 1)(z - 6)$ — real linear factors, as the theorem on self-adjoint operators predicts. The eigenvalues are $1$ and $6$ .

Eigenvectors. $(A - I)v = 0$ reads $v_1 + 2v_2 = 0$ , so $(2, -1)$ ; $(A - 6I)v = 0$ reads $2v_1 - v_2 = 0$ , so $(1, 2)$ . Check: $(2, -1) \cdot (1, 2) = 2 - 2 = 0$ .

Normalize. $\|(2, -1)\| = \|(1, 2)\| = \sqrt{5}$ , so

e_1 = \tfrac{1}{\sqrt{5}}(2, -1), \qquad e_2 = \tfrac{1}{\sqrt{5}}(1, 2), \qquad \mathcal{M}\big(T, (e_1, e_2)\big) = \begin{pmatrix} 1 & 0 \\ 0 & 6 \end{pmatrix}.

Problem 2 · Diagonalizable Is Not Enough

Given: each matrix is $\mathcal{M}(T)$ for an operator $T$ on $\mathbf{R}^2$ with respect to the standard basis (Euclidean inner product). For which $T$ does $\mathbf{R}^2$ have an orthonormal basis of eigenvectors of $T$ ?

✅ Correct! The matrix equals its transpose and the standard basis is orthonormal, so

T = T^*

; indeed

\tfrac{1}{\sqrt{2}}(1, 1)

(eigenvalue

3

) and

\tfrac{1}{\sqrt{2}}(1, -1)

(eigenvalue

-1

) form an orthonormal eigenbasis.

❌ Not quite. This

T

is diagonalizable —

T(1, 0) = (2, 0)

and

T(1, 2) = (4, 8)

give a basis of eigenvectors, for the distinct eigenvalues

2

and

4

— but

(1, 0) \cdot (1, 2) = 1 \neq 0

, and each eigenspace is a single line, so no other choice of eigenvectors is orthogonal. The matrix is not equal to its transpose, so

T \neq T^*

, and the real spectral theorem rules out every orthonormal eigenbasis.

❌ Not quite. Here

\mathcal{M}(T)^2 = 4\mathcal{M}(T) - 5I

, so the minimal polynomial is

z^2 - 4z + 5

, a quadratic with

b^2 = 16 < 20 = 4c

. It has no real zero, so

T

has no eigenvalue at all — it is

2I

plus a quarter turn (a rotation scaled by

\sqrt{5}

), not self-adjoint.

❌ Not quite. The only eigenvalue is

1

, and

T(x, y) = (x + y, y) = (x, y)

forces

y = 0

: every eigenvector lies on the line through

(1, 0)

, so there is not even a basis of eigenvectors, orthonormal or not.

Show solution

Over $\mathbf{R}$ , an orthonormal eigenbasis exists exactly when $T = T^*$ , and with respect to the (orthonormal) standard basis that means $\mathcal{M}(T)^{t} = \mathcal{M}(T)$ . Only (d) is symmetric.

(d): $T(1, 1) = (3, 3)$ and $T(1, -1) = (-1, 1)$ , with $(1, 1) \cdot (1, -1) = 0$ ; dividing by $\sqrt{2}$ gives the orthonormal eigenbasis.

(a) has a basis of eigenvectors $(1, 0)$ , $(1, 2)$ (eigenvalues $2$ , $4$ ) with $(1, 0) \cdot (1, 2) = 1 \neq 0$ — diagonalizable but not self-adjoint. (b) has minimal polynomial $z^2 - 4z + 5$ with $(-4)^2 < 4 \cdot 5$ , so no eigenvalue. (c) has eigenvectors only along $(1, 0)$ .

Problem 3 · A Repeated Eigenvalue in $\mathbf{R}^3$

Given: $T \in \mathcal{L}(\mathbf{R}^3)$ has matrix $B = \begin{pmatrix} 3 & 0 & 1 \\ 0 & 2 & 0 \\ 1 & 0 & 3 \end{pmatrix}$ with respect to the standard basis (Euclidean inner product). Find an orthonormal eigenbasis and the diagonal matrix of $T$ in it.

Which list is an orthonormal basis of eigenvectors of $T$ ?

With respect to the basis in (a), in that order, $\mathcal{M}(T)$ is

✅ Correct!

T(1, 0, 1) = (4, 0, 4)

T(0, 1, 0) = (0, 2, 0)

and

T(1, 0, -1) = (2, 0, -2)

; the three vectors are pairwise orthogonal, so the normalized list is an orthonormal eigenbasis with

\mathcal{M}(T) = \operatorname{diag}(4, 2, 2)

❌ Not quite. All three are eigenvectors —

T(1, 1, -1) = (2, 2, -2)

— but

(0, 1, 0) \cdot (1, 1, -1) = 1 \neq 0

. Eigenvectors for distinct eigenvalues are automatically orthogonal; two from the same eigenspace (eigenvalue

2

here) are not, so they must be chosen orthogonal.

❌ Not quite.

T(1, 0, 0) = (3, 0, 1)

is not a multiple of

(1, 0, 0)

, and likewise for

(0, 0, 1)

B

is not diagonal, so only

(0, 1, 0)

in this list is an eigenvector.

❌ Not quite.

T(0, 0, 1) = (1, 0, 3)

is not a multiple of

(0, 0, 1)

, and

(0, 0, 1)

is not orthogonal to

(1, 0, -1)

either; the missing eigenvector is

(1, 0, 1)

, for eigenvalue

4

❌ Not quite. Check each vector:

Bv

must be a multiple of

v

, and every pair must have dot product

0

❌ Not quite. These are the right eigenvalues in the wrong order. The

k

-th diagonal entry is the eigenvalue of the

k

-th basis vector, and the first vector

\tfrac{1}{\sqrt{2}}(1, 0, 1)

has

T(1, 0, 1) = (4, 0, 4)

: eigenvalue

4

comes first.

❌ Not quite. These are the diagonal entries of

B

itself, the matrix with respect to the standard basis. Changing the basis changes the diagonal:

T(1, 0, 1) = (4, 0, 4)

, so the first entry is

4

, not

3

❌ Not quite.

T(1, 0, -1) = (2, 0, -2) = 2(1, 0, -1)

: the minus sign in the third coordinate belongs to the vector, so the eigenvalue is

2

, not

-2

❌ Not quite. Apply

B

to each basis vector in order and read off the multiple.

Show solution

$B = B^{t}$ , so $T = T^*$ and an orthonormal eigenbasis exists.

Minimal polynomial. $(B - 2I)(B - 4I) = \begin{pmatrix} 1 & 0 & 1 \\ 0 & 0 & 0 \\ 1 & 0 & 1 \end{pmatrix}\begin{pmatrix} -1 & 0 & 1 \\ 0 & -2 & 0 \\ 1 & 0 & -1 \end{pmatrix} = 0$ , and $B$ is not a multiple of $I$ , so the minimal polynomial is $(z - 2)(z - 4)$ : eigenvalues $2$ and $4$ .

Eigenvalue 4: $(B - 4I)v = 0$ gives $v_2 = 0$ , $v_1 = v_3$ , so $(1, 0, 1)$ .

Eigenvalue 2: $(B - 2I)v = 0$ gives only $v_1 + v_3 = 0$ — a plane. Choose an orthogonal pair in it: $(0, 1, 0)$ and $(1, 0, -1)$ .

Pairwise dot products are $0$ ; the lengths are $\sqrt{2}$ , $1$ , $\sqrt{2}$ . So

e_1 = \tfrac{1}{\sqrt{2}}(1, 0, 1),\quad e_2 = (0, 1, 0),\quad e_3 = \tfrac{1}{\sqrt{2}}(1, 0, -1),\qquad \mathcal{M}\big(T, (e_1, e_2, e_3)\big) = \begin{pmatrix} 4 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2 \end{pmatrix}.

Problem 4 · Which Quadratic Is Always Invertible?

Given: $T$ is a self-adjoint operator on a finite-dimensional real inner product space $V$ — any such $V$ , any such $T$ . Which operator is guaranteed to be invertible?

✅ Correct!

b^2 = 4 < 20 = 4c

. Directly:

\langle (T^2 + 2T + 5I)v, v \rangle = \|Tv + v\|^2 + 4\|v\|^2 > 0

for

v \neq 0

❌ Not quite. Here

b^2 = 16 = 4c

, equality rather than

<

T^2 - 4T + 4I = (T - 2I)^2

, which is the zero operator when

T = 2I

. The theorem needs the strict inequality.

❌ Not quite.

b^2 = 25 > 24 = 4c

, so the quadratic has real zeros:

T^2 - 5T + 6I = (T - 2I)(T - 3I)

sends every eigenvector with eigenvalue

2

0

— for instance every vector when

T = 2I

❌ Not quite. Positive coefficients do not help:

b^2 = 16 > 12 = 4c

, and

T^2 + 4T + 3I = (T + I)(T + 3I)

is the zero operator for the self-adjoint

T = -I

Show solution

The theorem on invertible quadratic expressions applies exactly when $b^2 < 4c$ : only (b) qualifies ( $4 < 20$ ); (a) has $16 = 16$ , (c) has $25 > 24$ , (d) has $16 > 12$ .

Direct check for (b). Since $T = T^*$ , $\langle T^2 v, v \rangle = \langle Tv, Tv \rangle = \|Tv\|^2$ , and over $\mathbf{R}$ , $\|Tv + v\|^2 = \|Tv\|^2 + 2\langle Tv, v \rangle + \|v\|^2$ . Hence

\langle (T^2 + 2T + 5I)v, v \rangle = \|Tv\|^2 + 2\langle Tv, v \rangle + 5\|v\|^2 = \|Tv + v\|^2 + 4\|v\|^2 \ge 4\|v\|^2 > 0

for $v \neq 0$ , so $(T^2 + 2T + 5I)v \neq 0$ : the operator is injective, hence invertible since $V$ is finite-dimensional.

The others fail for $T = 2I$ , $T = 2I$ and $T = -I$ respectively, each a self-adjoint operator.

Solved: 0 / 4

Real Spectral Theorem: Self-Adjoint ⇔ Orthonormal Eigenbasis

IKey Formulas

IIWhy T2+bT+cIT^2 + bT + cIT2+bT+cI Cannot Send v≠0v \neq 0v=0 to 000

IIISymmetric Matrix, Perpendicular Eigenvectors

IV(1) ⇒ (2): Upper-Triangular and Self-Adjoint Leaves Only the Diagonal

VQuiz Questions

IIWhy $T^2 + bT + cI$ Cannot Send $v \neq 0$ to $0$