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5.3 Diagonalization (part 2)

Example

Diagonalize A = \(\begin{bmatrix} 5 & -3 & 0 & 9 \\ 0 & 3 & 1 & -2 \\ 0 & 0 & 2 & 0 \\ 0 & 0 & 0 & 2 \end{bmatrix}\)

Triangular, so eigenvalues are on the main diagonal:

\(\lambda = 5, 3, 2, 2\)

Find eigenvector for \(\lambda = 5\)

Solve \((A - \lambda I)\vec{x} = \vec{0}\)

\[\begin{bmatrix} 0 & -3 & 0 & 9 & 0 \\ 0 & -2 & 1 & -2 & 0 \\ 0 & 0 & -3 & 0 & 0 \\ 0 & 0 & 0 & -3 & 0 \end{bmatrix}\]

\(x_2 = x_3 = x_4 = 0\)

\(x_1\) free, choose \(x_1 = 1\)

so eigenvector \(\vec{v} = \begin{bmatrix} 1 \\ 0 \\ 0 \\ 0 \end{bmatrix}, \lambda = 5\)

\(\lambda = 3\)

\[\begin{bmatrix} 2 & -3 & 0 & 9 & 0 \\ 0 & 0 & 1 & -2 & 0 \\ 0 & 0 & -1 & 0 & 0 \\ 0 & 0 & 0 & -1 & 0 \end{bmatrix}\]

\(x_3 = x_4 = 0\)

\(x_2\) free

\(x_1 = \frac{3}{2} x_2\)

choose \(x_2 = 2\)

\(\vec{v} = \begin{bmatrix} 3 \\ 2 \\ 0 \\ 0 \end{bmatrix}, \lambda = 3\)

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\(\lambda = 2\)

\[\begin{bmatrix} 3 & -3 & 0 & 9 & 0 \\ 0 & 1 & 1 & -2 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \end{bmatrix}\]

\(x_3, x_4\) free

\(x_2 = -x_3 + 2x_4\)

\(x_1 = x_2 - 3x_4\)

\(= -x_3 + 2x_4 - 3x_4 = -x_3 - x_4\)

\(\vec{x} = x_3 \begin{bmatrix} -1 \\ -1 \\ 1 \\ 0 \end{bmatrix} + x_4 \begin{bmatrix} -1 \\ 2 \\ 0 \\ 1 \end{bmatrix}\)

eigenspace has dimension of 2, \(\lambda = 2\) has algebraic multiplicity of 2.

\(A = PDP^{-1}\)

\[P = \begin{bmatrix} 1 & 3 & -1 & -1 \\ 0 & 2 & -1 & 2 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix}\]
\[D = \begin{bmatrix} 5 & 0 & 0 & 0 \\ 0 & 3 & 0 & 0 \\ 0 & 0 & 2 & 0 \\ 0 & 0 & 0 & 2 \end{bmatrix}\]

Note:

This factorization is NOT unique because we can order columns of P however we want, as long as the corresponding eigenvalues are in the same order in D.

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the same columns of \( D \). Also, we can scale columns of \( P \) however we want.

A matrix is NOT diagonalizable if at least one \( \lambda \) is repeated AND that \( \lambda \)'s eigenspace does not have enough dimensions.

Simple example of non diagonalizable matrix

\[ A = \begin{bmatrix} 1 & 0 \\ 1 & 1 \end{bmatrix} \quad \lambda = 1, 1 \]

eigenvector for \( \lambda = 1 \):

\[ (A - \lambda I) \vec{x} = \vec{0} \quad \begin{bmatrix} 0 & 0 & 0 \\ 1 & 0 & 0 \end{bmatrix} \]
  • \( x_2 \) is free
  • \( x_1 = 0 \)
  • eigenspace dimension = 1
  • \( \lambda = 1 \) has multiplicity of 2

the only eigenvector is \( \vec{v} = \begin{bmatrix} 0 \\ 1 \end{bmatrix} \)

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If \( A \) is diagonalizable, it may or may not be invertible, because if at least one \( \lambda = 0 \), then \( A^{-1} \) does not exist.

What about the other way around? Does invertibility imply diagonalizability? No, see \( A = \begin{bmatrix} 1 & 0 \\ 1 & 1 \end{bmatrix} \)

If \( A \) is diagonalizable and invertible, is \( A^{-1} \) diagonalizable?

if \( A \) is diagonalizable, then \( A = PDP^{-1} \) (where \( D \) is a diagonal matrix)

if \( A \) is invertible, then \( D \) is invertible

\[ A = PDP^{-1} \]\[ A^{-1} = (PDP^{-1})^{-1} \]\[ = (P^{-1})^{-1} D^{-1} P^{-1} \]\[ = PD^{-1}P^{-1} \]

recall \( (AB)^{-1} = B^{-1}A^{-1} \)

is \( D^{-1} \) diagonal?

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If

\[ D = \begin{bmatrix} a & 0 & 0 \\ 0 & b & 0 \\ 0 & 0 & c \end{bmatrix} \]

then

\[ \left[ \begin{array}{ccc|ccc} a & 0 & 0 & 1 & 0 & 0 \\ 0 & b & 0 & 0 & 1 & 0 \\ 0 & 0 & c & 0 & 0 & 1 \end{array} \right] \sim \left[ \begin{array}{ccc|ccc} 1 & 0 & 0 & 1/a & 0 & 0 \\ 0 & 1 & 0 & 0 & 1/b & 0 \\ 0 & 0 & 1 & 0 & 0 & 1/c \end{array} \right] \]

\( D^{-1} \)

so \( D^{-1} \) is diagonal, and \( A^{-1} = P D^{-1} P^{-1} \)

so, \( A^{-1} \) is diagonalizable.

Example: 5x5 Matrix

If \( A \) is \( 5 \times 5 \) with two different eigenvalues. One eigenspace is 3-dimensional and the other eigenspace is 2-dimensional. Is \( A \) diagonalizable?

\[ \lambda = a, a, b, b, b \]

Eigenspace for \( a \):

  • is 2-D
  • two eigenvectors

Eigenspace for \( b \):

  • is 3-D
  • three eigenvectors

All linearly independent. So, yes, \( P \) matrix exists.

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Example: 4x4 Matrix

If \( A \) is \( 4 \times 4 \) w/ 3 different eigenvalues, one eigenspace is 1-dimensional and one of the other is two-dimensional. Is it possible that \( A \) is NOT diagonalizable?

\[ \lambda = a, b, c, c \]

For \( a \):

Eigenspace 1-D

1 vector

For \( c \):

Eigenspace is 2-D

2 vectors

\( b \) is an eigenvalue, so it has an eigenvector. It must have eigenspace that is 1-D and it is linearly independent from others because \( b \) is distinct from \( a \) and \( c \).

No, \( A \) is always diagonalizable