Proof of the real spectral theorem

Orthogonal bases and Gram-Schmidt

Proposition: (The Gram-Schmidt process) Let $V$ be a real inner product space of dimension $n$, and let $v_1,\dots ,v_k$ be a linearly independent set in $V$. Then there is a set $w_1,\dots ,w_k$ of vectors such that

• $⟨w_i,w_j⟩=0$ if $i\ne j$.
• the span of $w_1,\dots ,w_l$ is the same as the span of $v_1,\dots ,v_l$ for $l\le k$.

Proof: Let $w_1=v_1$ and $w_2=v_2-\frac{⟨v_2,w_1⟩}{{\Vert w_1\Vert }^2}{w}_1.$ Then the span of $w_2$ and $w_1$ is the same as that of $v_2$ and $v_1$, and $⟨w_2,w_1⟩=0$ by construction. Now suppose we have constructed $w_1,\dots ,w_l$ with the desired property. Set

$$w_{l+1}=v_{l+1}-\sum _{i=1}^l\frac{⟨v_{l+1},w_i⟩}{{\Vert w_i\Vert }^2}{w}_i.$$

Then $⟨w_{l+1},w_i⟩=0$ and the span property is preserved.

Orthogonal complements

If $W$ is a subspace of $V$, define $W^{\perp }=\{v:⟨v,w⟩=0\}$.

Proposition: $W^{\perp }$ is a subspace of $V$. Furthermore:

• $\dim W+\dim W^{\perp }=\dim V$ (so $W\cap W^{\perp }=0$.)
• $(W^{\perp }{)}^{\perp }=W$.
• if $U\subset W$, then $W^{\perp }\subset U^{\perp }$.

Proof: Suppose $\dim W=k$ and $\dim V=n$. Use Gram-Schmidt to construct an orthogonal basis $v_1,\dots ,v_n$ for $V$ whose first $k$ elements are an orthogonal basis for $W$. A vector

$$v=\sum a_i{v}_i$$

is in $W^{\perp }$ if and only if $a_i=0$ for $i=1,\dots ,k$.

Proposition: Suppose that $A$ is a self adjoint operator and $AW\subset W$. Then $A{W}^{\perp }\subset W^{\perp }$.

Proof: Suppose $z\in W^{\perp }$ and $w\in W$. Then

$$⟨Az,w⟩=⟨z,A^{\ast }w⟩=⟨z,Aw⟩=0$$

since $Aw\in W$.

Proof of the (real) spectral theorem

We have a self-adjoint map $A:V\to V$. Pick a basis for $V$ and use Gram-Schmidt to construct an orthonormal basis (an orthogonal basis where the elements all have norm $1$).

The $Q$-matrix for this basis is the identity, and so the inner product is just the dot product.

If $[A]$ is the matrix representation of $A$ in this basis, then the matrix representation of $A^{\ast }$ is the transpose of $[A]$. So since $A$ is self-adjoint, $[A]$ is symmetric.

We know that a symmetric matrix has a real eigenvalue ${\lambda }_1$ with eigenvector $v_1$.

Let $V_1$ be the orthogonal complement to the one-dimensional space $W$ spanned by $v_1$. Since $v_1$ is an eigenvector, $AW\subset W$. Therefore $A{V}_1\subset V_1$. Furthermore, if $x,y\in V_1$, then

$$⟨Ax,y⟩=⟨x,Ay⟩$$

so $A$ is self-adjoint as a linear map from $V_1$ to itself. Thus we can continue by induction to construct an orthogonal basis of eigenvectors for $A$.

Orthogonal matrices

Let $Q\in M_n(\mathbb{R})$ be a symmetric matrix. As such it is a self adjoint map from ${\mathbb{R}}^n$ to itself with respect to the usual dot product. Therefore there is a basis $v_1,\dots ,v_n$ of ${\mathbb{R}}^n$ consisting of orthonormal eigenvectors for the dot product $Q$- eigenvalues ${\lambda }_1,\dots ,{\lambda }_n$.

Let $P$ be the matrix whose columns are the vectors $v_i$ written in the standard basis of ${\mathbb{R}}^n$. Since the $v_i$ are an orthonormal basis, the matrix $P$ satisfies $P^{T}P=I$.

At the same time,

$$QP=P\mathrm{\Lambda }$$

where $\mathrm{\Lambda }$ is the diagonal matrix with entries ${\lambda }_i$. Since the $v_i$ are linearly independent, the matrix $P$ is invertible and $Q$ is diagonalizable:

$$P^{-1}QP=\mathrm{\Lambda }$$

The bilinear map $⟨v,w⟩$ defined by

$$⟨v,w⟩=v^{T}Qw$$

is an inner product provided that $⟨v,v⟩\ge 0$ with equality only one $v=0$. If we write $v\ne 0$ in terms of the orthogonal basis $v_1,\dots ,v_n$:

$$v=\sum a_i{v}_i$$

then we get

$$⟨v,v⟩=(\sum a_i{v}_i^T)Q(\sum a_i{v}_i)=\sum a_i{v}_i^T{\lambda }_i{v}_i=\sum a_i^2{\lambda }_i{v}_i^T{v}_i$$

which will be positive provided that all ${\lambda }_i>0$.

Example

Let $Q$ be the symmetric matrix

$$Q=\left(\begin{array}{cc}3& 1\\ 1& 3\end{array}\right)$$

Its eigenvalues are $2$ and $4$ with eigenvectors

$$\left[\begin{array}{c}-\sqrt{2}\\ \sqrt{2}\end{array}\right]$$

and

$$\left[\begin{array}{c}2\sqrt{2}\\ 2\sqrt{2}\end{array}\right].$$

The norm of a vector in the inner product given by $Q$ is

$$\Vert (x,y)\Vert =3x^2+2xy+3y^2$$

The level curves of this are ellipses, and the eigenvectors point in the directions of the major and minor axes of the ellipse.

These ellipses are the family

$$2(x-y{)}^2+4(x+y{)}^2=C$$

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