Complex Spectral Theorem

Andrew Thangaraj

Aug-Nov 2020


  • Vector space \(V\) over a scalar field \(F\)
    • \(F\): real field \(\mathbb{R}\) or complex field \(\mathbb{C}\) in this course
  • \(m\times n\) matrix A represents a linear map \(T:F^n\to F^m\)
    • dim null \(T+\) dim range \(T=\) dim \(V\)
    • Solution to \(Ax=b\) (if it exists): \(u+\) null\((A)\)
  • Four fundamental subspaces of a matrix
    • Column space, row space, null space, left null space
  • Eigenvalue \(\lambda\) and Eigenvector \(v\): \(Tv=\lambda v\)
    • There is a basis w.r.t. which a linear map is upper-triangular
    • If there is a basis of eigenvectors, linear map is diagonal w.r.t. it
  • Inner products, norms, orthogonality and orthonormal basis
    • There is an orthonormal basis w.r.t. which a linear map is upper-triangular
    • Orthogonal projection: distance from a subspace
  • Adjoint of a linear map: \(\langle Tv,w\rangle=\langle v,T^*w\rangle\)
    • null \(T=\) \((\)range \(T^*)^{\perp}\)
  • Self-adjoint: \(T=T^*\), Normal: \(TT^*=T^*T\)
    • Eigenvectors corresponding to different eigenvalues are orthogonal

Diagonalizable operators

\(T:V\to V\), diagonalizable

Eigenvector basis of \(V\): \(\{v_1,\ldots,v_n\}\)


Matrix of \(T\) w.r.t. eigenvector basis

\(D=\begin{bmatrix} \lambda_1&0&\cdots&0\\ 0&\lambda_2&\cdots&0\\ \vdots&\vdots&\ddots&\vdots\\ 0&0&\cdots&\lambda_n \end{bmatrix}\) (diagonal)

Matrices of diagonalizable operators

\(A\): \(n\times n\) matrix, represents \(T\) in standard basis

Matrix with \(v_i\) (coordinates in standard basis) as \(i\)-th column
\(S=\begin{bmatrix} \vdots&\cdots&\vdots&\cdots&\vdots\\ v_1&\cdots&v_i&\cdots&v_n\\ \vdots&\cdots&\vdots&\cdots&\vdots \end{bmatrix}\)

Let \(S^{-1}=\begin{bmatrix} \vdots&\vdots&\vdots\\ \cdots&u^T_i&\cdots\\ \vdots&\vdots&\vdots \end{bmatrix}\), matrix with \(u^T_i\) as \(i\)-th row

\(A=SDS^{-1}=\begin{bmatrix} \vdots&\cdots&\vdots&\cdots&\vdots\\ v_1&\cdots&v_i&\cdots&v_n\\ \vdots&\cdots&\vdots&\cdots&\vdots \end{bmatrix}\begin{bmatrix} \lambda_1&0&\cdots&0\\ 0&\lambda_2&\cdots&0\\ \vdots&\vdots&\ddots&\vdots\\ 0&0&\cdots&\lambda_n \end{bmatrix}\begin{bmatrix} \vdots&\vdots&\vdots\\ \cdots&u^T_i&\cdots\\ \vdots&\vdots&\vdots \end{bmatrix}\)


General form of matrices of diagonalizable operators

\(T:V\to V\): diagonalizable

There exists a basis s.t. matrix of \(T\) has the following form:

\(v_i\): eigenvectors in the basis, \(u_i\): rows of inverse matrix

Suppose rank \(T=r\).

There exists a basis s.t. matrix of \(T\) has the following form:

\(\lambda_i\): nonzero eigenvalues with corresponding eigenvectors \(v_i\), \(1\le i\le r\)


  1. \(A=\begin{bmatrix} 1&0\\ 0&2 \end{bmatrix}\) (self-adjoint, diagonal)

    \(A=\begin{bmatrix} 1\\ 0 \end{bmatrix}\begin{bmatrix} 1&0 \end{bmatrix}+2\begin{bmatrix} 0\\ 1 \end{bmatrix}\begin{bmatrix} 0&1 \end{bmatrix}\)

  2. \(A=\begin{bmatrix} 1&2\\ 3&2 \end{bmatrix}\) (not self-adjoint)


    \(S=\begin{bmatrix} 1&2\\ -1&3 \end{bmatrix}\), \(S^{-1}=\begin{bmatrix} 3/5&-2/5\\ 1/5&1/5 \end{bmatrix}\)

    \(A=-\begin{bmatrix} 1\\ -1 \end{bmatrix}\begin{bmatrix} 3/5&-2/5 \end{bmatrix}+4\begin{bmatrix} 2\\ 3 \end{bmatrix}\begin{bmatrix} 1/5&1/5 \end{bmatrix}\)

Example: Normal operator

\(A=\begin{bmatrix} 2&-3\\ 3&2 \end{bmatrix}\) (normal)


\(S=\frac{1}{\sqrt{2}}\begin{bmatrix} i&-i\\ 1&1 \end{bmatrix}\), \(S^{-1}=\frac{1}{\sqrt{2}}\begin{bmatrix} -i&1\\ i&1 \end{bmatrix}\)

\(A=\frac{2+3i}{2}\begin{bmatrix} i\\ 1 \end{bmatrix}\begin{bmatrix} -i&1 \end{bmatrix}+\frac{2-3i}{2}\begin{bmatrix} -i\\ 1 \end{bmatrix}\begin{bmatrix} i&1 \end{bmatrix}\)


Eigenvector basis: Orthonormal

Complex spectral theorem

\(V\): inner product space over \(C\), \(T:V\to V\)

The following are equivalent.

  1. \(T\) is normal.
  2. \(V\) has an orthonormal basis of eigenvectors of \(T\).
  3. \(T\) is diagonal w.r.t. an orthonormal basis.


Clearly, (2) implies (3), and (3) implies (2)

Proof of (3) implies (1)

\(B\): orthonormal basis s.t. \(M(T,B)\) is diagonal

Then, \(M(T^*,B)\): conjugate-transpose of \(M(T,B)\), which is diagonal

So, \(M(T,B)M(T^*,B)=M(T^*,B)M(T,B)\), or \(T\) commutes with \(T^*\)

Proof (continued)

Proof of (1) implies (3)

Schur’s theorem: \(T\) is upper-triangular w.r.t. an orthonormal basis

Orthonormal basis: \(\{e_1,\cdots,e_n\}\)

\(M(T,B)=\begin{bmatrix} a_{11}&\cdots&a_{1n}\\ &\ddots&\vdots\\ 0 & &a_{nn} \end{bmatrix}\), \(M(T^*,B)=\begin{bmatrix} \overline{a_{11}}& &0\\ \vdots &\ddots& \\ \overline{a_{1n}}&\cdots&\overline{a_{nn}} \end{bmatrix}\)

We will show that \(M(T,B)\) is, in fact, diagonal

\(\lVert Te_1\rVert^2=\lvert a_{11}\rvert^2=\lVert T^*e_1\rVert^2=\lvert a_{11}\rvert^2+\cdots+\lvert a_{1n}\rvert^2\)

So, \(a_{12}=\cdots=a_{1n}=0\)

\(\lVert Te_2\rVert^2=\lvert a_{22}\rvert^2=\lVert T^*e_2\rVert^2=\lvert a_{22}\rvert^2+\cdots+\lvert a_{2n}\rvert^2\)

So, \(a_{23}=\cdots=a_{2n}=0\)

and so on…

Matrices of normal operators

\(A\): \(n\times n\) matrix, represents a normal operator \(T\) in standard basis

\(T\) is diagonal w.r.t. orthonormal eigenvector basis \(\{e_1,\cdots,e_n\}\)

Corresponding eigenvalues: \(\{\lambda_1,\ldots,\lambda_n\}\)

Matrix with \(e_i\) (coordinates in standard basis) as \(i\)-th column
\(S=\begin{bmatrix} \vdots&\cdots&\vdots&\cdots&\vdots\\ e_1&\cdots&e_i&\cdots&e_n\\ \vdots&\cdots&\vdots&\cdots&\vdots \end{bmatrix}\)

\(S^{-1}=\begin{bmatrix} \vdots&\vdots&\vdots\\ \cdots&\overline{e^T_i}&\cdots\\ \vdots&\vdots&\vdots \end{bmatrix}\), matrix with \(\overline{e^T_i}\) as \(i\)-th row

\(A=SDS^{-1}=\lambda_1e_1\overline{e^T_1}+\cdots+\lambda_n e_n \overline{e^T_n}\)