Andrew Thangaraj
Aug-Nov 2020
Concept | Real | Complex |
---|---|---|
Linear maps | - | - |
Fundamental spaces | rowspace \(=\) \((\)null\()^{\perp}\) | Need not hold |
Eigenvalues | Need not be real | At least one complex exists |
Inner product | Dot product | Conjugate dot product |
Example (\(A\): \(n\times n\) real matrix with real eigenvalue \(\lambda\))
Then, eigenvector \(\in\) null\((A-\lambda I)\) and is real
Example (normal, but not self-adjoint)
\(A=\begin{bmatrix} 0&1\\ -1&0 \end{bmatrix}\)
\(\lambda = i,-i\)
Eigenvectors: \((i,1)\), \((-i,1)\)
\(V\): vector space over \(\mathbb{R}\)
\(T:V\to V\), self-adjoint
Eigenvalues of \(T\) are real
\(A\): \(n\times n\) matrix representing \(T\)
Roots of det\((A-\lambda I)\) have to be all real.
What about eigenvectors?
What about geometric multiplicity?
What about diagonalizability?
\(V\): vector space over \(\mathbb{R}\)
\(T:V\to V\), an operator
The following are equivalent:
- \(T\) is self-adjoint
- \(V\) has an orthonormal basis of eigenvectors of \(T\)
- \(T\) is diagonal w.r.t. an orthonormal basis
Proof
Clearly, (2) implies (3) and vice versa
Proof of (3) implies (1)
Let \(D\) be the diagonal matrix representing \(T\) w.r.t. an orthonormal basis
Diagonal values of \(D\) are the real eigenvalues
So, conjugate-transpose of \(D\) is equal to \(D\), or \(T\): self-adjoint
Proof of (1) implies (3)
Let \(\lambda\) be a real eigenvalue of \(T\) with a real eigenvector \(v\).
Extend \(v\) to an orthonormal basis: \(\{v,u_1,\ldots,u_{n-1}\}\) (all real)
\(A=\begin{bmatrix} \lambda&a_{12}&\cdots&a_{1n}\\ 0 & & & \\ \vdots & & A_1 & \\ 0 & & & \end{bmatrix}\) (matrix of \(T\) w.r.t. above basis)
Since \(A=A^T\), we have the following:
\(a_{12}=\cdots=a_{1n}=0\)
\(A_1=A^T_1\): \((n-1)\times(n-1)\), self-adjoint
\(A_1\): represents a self-adjoint operator from \(\{u_1,\ldots,u_{n-1}\}\to\{u_1,\ldots,u_{n-1}\}\)
Repeat same argument with \(A_1\)
Finally, get a diagonal matrix for \(T\) w.r.t. an orthonormal basis for \(V\)
\(A\): \(n\times n\) real, symmetric matrix (representing a self-adjoint operator w.r.t. standard basis)
There is a real, orthonormal basis \(\{e_1,\ldots,e_n\}\) s.t.
\(e_i\) is an eigenvector of \(A\), or \(Ae_i=\lambda e_i\), \(\lambda_i\) real
\(A=\lambda_1e_1\overline{e^T_1}+\cdots+\lambda_ne_n\overline{e^T_n}\)
Example
\(A=\begin{bmatrix} 14&-13&8\\ -13&14&8\\ 8&8&-7 \end{bmatrix}\)
\(e_1=\frac{1}{\sqrt{2}}(1,-1,0)\), \(e_2=\frac{1}{\sqrt{3}}(1,1,1)\), \(e_3=\frac{1}{\sqrt{6}}(1,1,-2)\)
\(\lambda_1=27\), \(\lambda_2=9\), \(\lambda_3=-15\)
\(A=\frac{27}{2}\begin{bmatrix}1\\-1\\0\end{bmatrix}\begin{bmatrix}1&-1&0\end{bmatrix} +\frac{9}{3}\begin{bmatrix}1\\1\\1\end{bmatrix}\begin{bmatrix}1&1&1\end{bmatrix} -\frac{15}{6}\begin{bmatrix}1\\1\\-2\end{bmatrix}\begin{bmatrix}1&1&-2\end{bmatrix}\)
\(\{e_1,\ldots,e_n\}\): orthonormal basis
\(A=\lambda_1e_1\overline{e^T_1}+\cdots+\lambda_ne_n\overline{e^T_n}\)
Type | \(e_i\) | \(\lambda_i\) |
---|---|---|
Normal (complex) | complex | complex |
Self-adjoint (complex) | complex | real |
Self-adjoint (real) | real | real |
Powers of \(A\)
\(A^k=\lambda^k_1e_1\overline{e^T_1}+\cdots+\lambda^k_ne_n\overline{e^T_n}\)
Order eigenvalues by magnitude \(\lvert\lambda_1\rvert \ge \cdots \ge \lvert\lambda_n\rvert\)
\(A^k\to\lambda^k_1e_1\overline{e^T_1}\) as \(k\to\infty\) (assume \(|\lambda_1|\) is the unique maximum)
Rank-\(r\) approximation of \(A\): \(\lambda_1e_1\overline{e^T_1}+\cdots+\lambda_re_r\overline{e^T_r}\)