Similar Matrices
Two matrices A and B are similar if there exists invertible P such that A = P B P − 1 \text{Two matrices A and B are similar if there exists invertible P such that } A = PBP^{-1} Two matrices A and B are similar if there exists invertible P such that A = PB P − 1 : If A and B are similar, then they have the same eigenvalues and determinants
Proof
λ I n − A = λ I n − P B P − 1 = λ P I n P − 1 − P B P − 1 = P ( λ I n P − 1 − B P − 1 ) = P ( λ I n − B ) P − 1 \lambda I_n - A = \lambda I_n - PBP^{-1} = \lambda PI_nP^{-1} - PBP^{-1} = P(\lambda I_nP^{-1} - BP^{-1}) \\
= P(\lambda I_n - B)P^{-1} λ I n − A = λ I n − PB P − 1 = λ P I n P − 1 − PB P − 1 = P ( λ I n P − 1 − B P − 1 ) = P ( λ I n − B ) P − 1 Upon taking determinants, we get that the characteristic polynomial of A is equal to the characteristic polynomial of B. Notated as PA (λ) = PB (λ).
v → is an eigenvector for A ⟹ P − 1 v → is an eigenvector of B u → is an eigenvector for B ⟹ P u → is an eigenvector of A \overrightarrow{v} \text{ is an eigenvector for A } \implies P^{-1}\overrightarrow{v} \text{ is an eigenvector of B} \\
\overrightarrow{u} \text{ is an eigenvector for B } \implies P\overrightarrow{u} \text{ is an eigenvector of A} \\ v is an eigenvector for A ⟹ P − 1 v is an eigenvector of B u is an eigenvector for B ⟹ P u is an eigenvector of A Example
A = ( 1 2 3 2 3 2 1 2 ) , B = ( 2 0 0 − 1 ) , P = ( 1 1 1 − 1 ) ( 1 1 1 − 1 ) ( 2 0 0 − 1 ) ∗ 1 2 ( 1 1 1 − 1 ) = 1 2 ( 1 1 1 − 1 ) ( 2 2 − 1 1 ) = 1 2 ( 1 3 3 1 ) A = \begin{pmatrix}
\frac{1}{2} & \frac{3}{2} \\
\frac{3}{2} & \frac{1}{2}
\end{pmatrix},
B = \begin{pmatrix}
2 & 0 \\
0 & -1
\end{pmatrix},
P = \begin{pmatrix}
1 & 1 \\
1 & -1
\end{pmatrix} \\
\begin{pmatrix}
1 & 1 \\
1 & -1
\end{pmatrix}
\begin{pmatrix}
2 & 0 \\
0 & -1
\end{pmatrix}
* \frac{1}{2}
\begin{pmatrix}
1 & 1 \\
1 & -1
\end{pmatrix} \\
= \frac{1}{2}
\begin{pmatrix}
1 & 1 \\
1 & -1
\end{pmatrix}
\begin{pmatrix}
2 & 2 \\
-1 & 1
\end{pmatrix} =
\frac{1}{2}
\begin{pmatrix}
1 & 3 \\
3 & 1
\end{pmatrix} A = ( 2 1 2 3 2 3 2 1 ) , B = ( 2 0 0 − 1 ) , P = ( 1 1 1 − 1 ) ( 1 1 1 − 1 ) ( 2 0 0 − 1 ) ∗ 2 1 ( 1 1 1 − 1 ) = 2 1 ( 1 1 1 − 1 ) ( 2 − 1 2 1 ) = 2 1 ( 1 3 3 1 ) In this example, the eigenvalues of A and B are 2 and -1 (based on B).
( 1 0 ) is a 2-eigenvector of B (because you simply have to multiply by the top row) ( 0 − 1 ) is a -1-eigenvector of B \begin{pmatrix} 1 \\ 0 \end{pmatrix} \text{ is a 2-eigenvector of B (because you simply have to multiply by the top row)} \\
\begin{pmatrix} 0 \\ -1 \end{pmatrix} \text{ is a -1-eigenvector of B} \\ ( 1 0 ) is a 2-eigenvector of B (because you simply have to multiply by the top row) ( 0 − 1 ) is a -1-eigenvector of B ( 1 1 1 − 1 ) ( 1 0 ) = ( 1 1 ) is a 2-eigenvector for A ( 1 1 1 − 1 ) ( 0 1 ) = ( 1 − 1 ) is a -1-eigenvector for A \begin{pmatrix}
1 & 1 \\
1 & -1
\end{pmatrix}
\begin{pmatrix} 1 \\ 0 \end{pmatrix} =
\begin{pmatrix} 1 \\ 1 \end{pmatrix} \text{ is a 2-eigenvector for A} \\
\begin{pmatrix}
1 & 1 \\
1 & -1
\end{pmatrix}
\begin{pmatrix} 0 \\ 1 \end{pmatrix} =
\begin{pmatrix} 1 \\ -1 \end{pmatrix} \text{ is a -1-eigenvector for A} ( 1 1 1 − 1 ) ( 1 0 ) = ( 1 1 ) is a 2-eigenvector for A ( 1 1 1 − 1 ) ( 0 1 ) = ( 1 − 1 ) is a -1-eigenvector for A Diagonalization
A diagonal matrix is a square matrix whose only potentially nonzero entries lie on the diagonal. Note that you CAN have zeroes on the diagonal (thus, the zero matrix is a diagonal matrix), but you can NOT have nonzeroes off of the diagonal.
Example
We want to turn A = ( 1 6 5 2 ) into D = ( 7 0 0 − 4 ) to more easily find its eigenvalues. A matrix is diagonalizable if there is a diagonal matrix D such that A = P D P − 1 (i.e. A and D are similar). \text{We want to turn }
A = \begin{pmatrix}
1 & 6 \\
5 & 2
\end{pmatrix} \text{ into }
D = \begin{pmatrix}
7 & 0 \\
0 & -4
\end{pmatrix}
\text{ to more easily find its eigenvalues.} \\
\text{A matrix is diagonalizable if there is a diagonal matrix D such that } A = PDP^{-1} \text{ (i.e. A and D are similar).} We want to turn A = ( 1 5 6 2 ) into D = ( 7 0 0 − 4 ) to more easily find its eigenvalues. A matrix is diagonalizable if there is a diagonal matrix D such that A = P D P − 1 (i.e. A and D are similar). A = ( 1 6 5 2 ) , D = ( 7 0 0 − 4 ) Instead of using A = P D P − 1 we can use A P = P D From last time: 7 is an eigenvalue with eigenvector ( 1 1 ) and -4 is an eigenvalue with eigenvector ( − 6 5 ) P = ( 1 − 6 1 5 ) , d e t ( P ) = 11 A P = ( A ( 1 1 ) A ( − 6 5 ) ) = ( 7 24 7 − 20 ) P D = ( 1 − 6 1 5 ) ( 7 0 0 − 4 ) = ( 7 24 7 − 20 ) Thus, we have verified that AP = PD. A = \begin{pmatrix}
1 & 6 \\
5 & 2
\end{pmatrix},
D = \begin{pmatrix}
7 & 0 \\
0 & -4
\end{pmatrix} \\
\text{Instead of using } A = PDP^{-1} \text{ we can use } AP = PD \\
\text{From last time: 7 is an eigenvalue with eigenvector } \begin{pmatrix} 1 \\ 1 \end{pmatrix} \text{ and -4 is an eigenvalue with eigenvector } \begin{pmatrix} -6 \\ 5 \end{pmatrix} \\
P = \begin{pmatrix}
1 & -6 \\
1 & 5
\end{pmatrix}, det(P) = 11 \\
AP = (A\begin{pmatrix} 1 \\ 1 \end{pmatrix} A \begin{pmatrix} -6 \\ 5 \end{pmatrix}) =
\begin{pmatrix}
7 & 24 \\
7 & -20
\end{pmatrix} \\
PD = \begin{pmatrix}
1 & -6 \\
1 & 5
\end{pmatrix}
\begin{pmatrix}
7 & 0 \\
0 & -4
\end{pmatrix} =
\begin{pmatrix}
7 & 24 \\
7 & -20
\end{pmatrix} \\
\text{Thus, we have verified that AP = PD.} A = ( 1 5 6 2 ) , D = ( 7 0 0 − 4 ) Instead of using A = P D P − 1 we can use A P = P D From last time: 7 is an eigenvalue with eigenvector ( 1 1 ) and -4 is an eigenvalue with eigenvector ( − 6 5 ) P = ( 1 1 − 6 5 ) , d e t ( P ) = 11 A P = ( A ( 1 1 ) A ( − 6 5 ) ) = ( 7 7 24 − 20 ) P D = ( 1 1 − 6 5 ) ( 7 0 0 − 4 ) = ( 7 7 24 − 20 ) Thus, we have verified that AP = PD. If we know that eigenvalues of n x n matrix A are λ 1 , . . . , λ n with eigenvectors v 1 → , . . . , v n → D = ( λ 1 . . . 0 ⋮ ⋱ ⋮ 0 . . . λ n ) , P = ( v 1 → . . . v n → ) Only applies if A has n distinct, linearly independent eigenvectors. \text{If we know that eigenvalues of n x n matrix A are } \lambda _1, ..., \lambda _n \text{ with eigenvectors } \overrightarrow{v_1}, ..., \overrightarrow{v_n} \\
D = \begin{pmatrix}
\lambda _1 & ... & 0 \\
\vdots & \ddots & \vdots \\
0 & ... & \lambda _n
\end{pmatrix},
P = \begin{pmatrix} \overrightarrow{v_1} & ... & \overrightarrow{v_n} \end{pmatrix} \\
\text{Only applies if A has n distinct, linearly independent eigenvectors.} If we know that eigenvalues of n x n matrix A are λ 1 , ... , λ n with eigenvectors v 1 , ... , v n D = ⎝ ⎛ λ 1 ⋮ 0 ... ⋱ ... 0 ⋮ λ n ⎠ ⎞ , P = ( v 1 ... v n ) Only applies if A has n distinct, linearly independent eigenvectors.