Section 4.5 Similarity and Diagonalization
Definition: If [latex]A[/latex] and [latex]B[/latex] are [latex]n\times n[/latex] matrices, then [latex]A[/latex] is similar to [latex]B[/latex] if there is an invertible matrix [latex]P[/latex] such that [latex]P^{-1}AP=B[/latex] or [latex]A=PBP^{-1}[/latex]. We write [latex]A\sim B[/latex].
Remark: We can write [latex]Q=P^{-1}[/latex] because [latex]P[/latex] is invertible then we have [latex]QAQ^{-1}=B[/latex], i.e. [latex]Q^{-1}BQ=A[/latex] and [latex]B[/latex] is similar to [latex]A[/latex]. Changing [latex]A[/latex] into [latex]P^{-1}AP=B[/latex] is called a similarity transformation.
Example1: If [latex]A[/latex] is similar to [latex]B[/latex] and either [latex]A[/latex] or [latex]B[/latex] is diagonalizable, show that the other is also diagonalizable.
Exercise 1: Show that if [latex]A[/latex] is similar to [latex]B[/latex] then [latex]\mbox{det}A=\mbox{det}B[/latex].
Theorem: If matrices [latex]A[/latex] and [latex]B[/latex] are similar, then they have the same characteristic
polynomial and hence the same eigenvalues (with the same multiplicities).
Remark: 1. Unfortunately, the converse of the theorem is not true. For example [latex]A=\left[\begin{array}{cc} 1 & 2\\ 0 & 3 \end{array}\right][/latex] and [latex]B=\left[\begin{array}{cc} 1 & 0\\ 0 & 3 \end{array}\right][/latex] have the same characteristic polynomial but they are not similar.
2. Two matrices that are row equivalent do not mean they are similar to each other. For example [latex]B=EA[/latex] where [latex]E[/latex] is just elementary matrix, and it does not mean [latex]A[/latex] is similar to [latex]B[/latex]. Therefore, we cannot use row reduction to get the eigenvalues.
Theorem: If [latex]A[/latex] and [latex]B[/latex] are similar [latex]n\times n[/latex] matrices, then [latex]A[/latex] and [latex]B[/latex] have the same determinant, rank, characteristic polynomial, and eigenvalues.
Remark: 1. A square matrix [latex]A[/latex] is diagonalizable then there exists an invertible matrix [latex]P[/latex] such that [latex]P^{-1}AP=D[/latex] is a diagonal matrix, that is [latex]A[/latex] is similar to a diagonal matrix [latex]D[/latex].
2.The set of all solutions of [latex](A-\lambda I)\overrightarrow{x}=0[/latex] is just the null space of the matrix [latex]A-\lambda I[/latex]. So this set is a subspace of [latex]\mathbb{R}^{n}[/latex] and is called the eigenspace of [latex]A[/latex] corresponding to $\lambda$. The eigenspace consists of the zero vector and all the eigenvectors corresponding to [latex]\lambda[/latex].
3. If [latex]\overrightarrow{x}[/latex] is in the eigenspace of some eigenvalue [latex]\lambda[/latex], then [latex]A\overrightarrow{x}[/latex] is just a scalar of [latex]\overrightarrow{x}[/latex]. This is saying the transformation defined by [latex]A[/latex] transforms the eigenspace into scalar multiple of itself.
Theorem: If [latex]\overrightarrow{v_{1}},...,\overrightarrow{v_{p}}[/latex] are eigenvectors that correspond to distinct eigenvalues [latex]\lambda_{1},...,\lambda_{p}[/latex] of an matrix [latex]A[/latex], then the set [latex]\{\overrightarrow{v_{1}},...,\overrightarrow{v_{p}}\}[/latex] is linearly independent.
Theorem: An [latex]n\times n[/latex] matrix [latex]A[/latex] is diagonalizable if and only if [latex]A[/latex] has [latex]n[/latex] linearly independent eigenvectors.
Remark: [latex]A[/latex] is diagonalizable if and only if there are enough eigenvectors to form a basis of [latex]\mathbb{R}^{n}[/latex]. We call such a basis an eigenvector basis of [latex]\mathbb{R}^{n}[/latex].
Theorem: Let [latex]A[/latex] be a matrix whose distinct eigenvalues are [latex]\lambda_{1},....,\lambda_{p}[/latex].
a. For [latex]1\leq k\leq p[/latex], the dimension of the eigenspace for [latex]\lambda_{k}[/latex] is less than or equal to the
multiplicity of the eigenvalue [latex]\lambda_{k}[/latex].
b. The matrix [latex]A[/latex] is diagonalizable if and only if the sum of the dimensions of the eigenspaces equals [latex]n[/latex], and this happens if and only if (i) the characteristic polynomial factors completely into linear factors and (ii) the dimension of the eigenspace for each [latex]\lambda_{k}[/latex] equals the multiplicity of [latex]\lambda_{k}[/latex].
c. If [latex]A[/latex] is diagonalizable and [latex]B_{k}[/latex] is a basis for the eigenspace corresponding to [latex]\lambda_{k}[/latex]
for each [latex]k[/latex], then the total collection of vectors in the sets [latex]B_{1},...,B_{p}[/latex] forms an eigenvector basis for [latex]\mathbb{R}^{n}[/latex].
Example 2: Diagonalize the following matrix, if possible.
[latex]A=\left[\begin{array}{cccc} 1 & 1 & 4 & 0\\ 0 & 1 & 1 & 6\\ 0 & 0 & -2 & 1\\ 0 & 0 & 0 & -2 \end{array}\right][/latex]
Exercise 2: Diagonalize the following matrix, if possible.
[latex]A=\left[\begin{array}{cccc} 3 & 0 & 0 & 0\\ 1 & -1 & 0 & 0\\ 0 & 1 & 2 & 0\\ 1 & 0 & 2 & -1 \end{array}\right][/latex]
Example 3: Diagonalize the following matrix, if possible.
[latex]A=\left[\begin{array}{ccc} 3 & 0 & 6\\ 0 & -3 & 0\\ 5 & 0 & 2 \end{array}\right][/latex]
Exercise 3: Diagonalize the following matrix, if possible.
[latex]A=\left[\begin{array}{ccc} 4 & 0 & 0\\ 0 & 2 & 2\\ 0 & 3 & 1 \end{array}\right][/latex]
GroupWork 1: True or False. All matrices are [latex]n\times n[/latex] matrices.
a. [latex]A[/latex] is diagonalizable with eigenvalues [latex]\lambda_{1},...,\lambda_{n}[/latex] then [latex]\text{det}A=\lambda_{1}\cdots\lambda_{n}[/latex].
b. If [latex]\mathbb{R}^{n}[/latex] has a basis of eigenvectors of [latex]A[/latex], then [latex]A[/latex] is diagonalizable.
c. [latex]A[/latex] is diagonalizable if and only if [latex]A[/latex] has $n$ eigenvalues, counting multiplicities.
d. If [latex]A[/latex] is invertible, then [latex]A[/latex] is diagonalizable.
e. An eigenspace of [latex]A[/latex] is a null space of a certain matrix.
GroupWork 2: [latex]A[/latex] is a [latex]6\times6[/latex] matrix with two eigenvalues. One eigenspace is 4-dimensional, and the other eigenspace is 2-dimensional. Is [latex]A[/latex] diagonalizable? Why?
GroupWork 3: If [latex]A[/latex] is an [latex]n\times n[/latex] matrix with [latex]n[/latex] distinct eigenvalues, show [latex]A[/latex] is diagonalizable.
GroupWork 4: Show that if [latex]A[/latex] is diagonalizable then [latex]A[/latex] is similar to [latex]A^{T}[/latex].
GroupWork 5: True or False. All matrices are [latex]n\times n[/latex] matrices. Justify each answer.
a. [latex]A[/latex] is diagonalizable if [latex]A[/latex] has [latex]n[/latex] eigenvectors.
b. [latex]A[/latex] is diagonalizable if [latex]A[/latex] has [latex]n[/latex] distinct eigenvectors.
c. [latex]A\sim B[/latex], then [latex]\text{det}(A-xI)=\text{det}(B-xI)[/latex].
d. [latex]A[/latex] is diagonalizable, then [latex]A[/latex] is invertible.
GroupWork 6: Let [latex]A[/latex] be an [latex]3\times3[/latex] matrix with 2 eigenvalues. Each eigenspace is one-dimensional. Is [latex]A[/latex] diagonalizable?
Why?
GroupWork7: [latex]A[/latex] is a [latex]5\times5[/latex] matrix with [latex]3[/latex] eigenvalues. Two of the eigenspaces are 2-dimensional. Is it possible that [latex]A[/latex] is not diagonalizable?