Line 3: | Line 3: | ||

* if for some invertible $M$ we can write $B = M^{-1} A \, M$ | * if for some invertible $M$ we can write $B = M^{-1} A \, M$ | ||

+ | |||

+ | == Invariant Subspaces == | ||

+ | Invariant subspace | ||

+ | * we know that if $x$ is the [[Eigenvalues and Eigenvectors|eigenvector]] of $A$, then $Ax = \lambda x$ | ||

+ | * $A$ only "streches" $x$ - but it remains in the same 1-dimensional subspace | ||

+ | * so such subspace $S$ formed by $x$ is invariant: | ||

+ | * if $x \in S \Rightarrow Ax \in S$ | ||

+ | |||

+ | Invariant column space | ||

+ | * If there exists such $M$ that $AM = MB$ | ||

+ | * then the $\text{range}(A)$ (the [[Column Space]] of $A$) is invariant | ||

Line 13: | Line 24: | ||

* we may take another $M \ne \Lambda$ and will get another matrix similar to $A$ (not necessarily diagonal) | * we may take another $M \ne \Lambda$ and will get another matrix similar to $A$ (not necessarily diagonal) | ||

− | + | A ''family'' of similar matrices for $A$ is a set of all matrices similar for $A$ (for different $M$s) | |

− | A family of similar matrices for $A$ is a set of matrices similar for $A$ for different $M$ | + | |

− | + | ||

Line 31: | Line 40: | ||

* Let $\mathbf x^*$ be $M^{-1} \mathbf x$, so we have | * Let $\mathbf x^*$ be $M^{-1} \mathbf x$, so we have | ||

* $B \mathbf x^* = \lambda \mathbf x^*$ | * $B \mathbf x^* = \lambda \mathbf x^*$ | ||

− | * so matrices $A$ and $B$ share the same eigenvalue $\lambda$ | + | * so matrices $A$ and $B$ share the same eigenvalue $\lambda$ |

− | + | * but not the eigenvector! $\mathbf x \ne \mathbf x^*$ | |

+ | === Diagonalization === | ||

For diagonalization | For diagonalization | ||

* $A = S^{-1} \Lambda \, S$ eigenvalues stay the same, but eigenvectors become unit vectors | * $A = S^{-1} \Lambda \, S$ eigenvalues stay the same, but eigenvectors become unit vectors | ||

− | |||

What if for some $i \ne j$, $\lambda_i = \lambda_j$? | What if for some $i \ne j$, $\lambda_i = \lambda_j$? | ||

Line 64: | Line 73: | ||

Suppose $A$ has a family of similar matrices | Suppose $A$ has a family of similar matrices | ||

* then the Jordan Form is the most diagonal matrix of the family | * then the Jordan Form is the most diagonal matrix of the family | ||

+ | |||

== Sources == | == Sources == | ||

* [[Linear Algebra MIT 18.06 (OCW)]] | * [[Linear Algebra MIT 18.06 (OCW)]] | ||

+ | * [[Matrix Computations (book)]] | ||

[[Category:Linear Algebra]] | [[Category:Linear Algebra]] |

We say that two $n \times n$ matrices $A$ and $B$ are *similar*

- if for some invertible $M$ we can write $B = M^{-1} A \, M$

Invariant subspace

- we know that if $x$ is the eigenvector of $A$, then $Ax = \lambda x$
- $A$ only "streches" $x$ - but it remains in the same 1-dimensional subspace
- so such subspace $S$ formed by $x$ is invariant:
- if $x \in S \Rightarrow Ax \in S$

Invariant column space

- If there exists such $M$ that $AM = MB$
- then the $\text{range}(A)$ (the Column Space of $A$) is invariant

Suppose $A$ has all its eigenvalues

- then if we diagonalize $A$, we have $S^{-1} A \, S = \Lambda$
- so $A$ is similar to $\Lambda$
- Here $M = \Lambda$
- we may take another $M \ne \Lambda$ and will get another matrix similar to $A$ (not necessarily diagonal)

A *family* of similar matrices for $A$ is a set of all matrices similar for $A$ (for different $M$s)

What is similar about such $A$ and $B$?

- they have the same eigenvalues!

Let's check it:

- let $A \mathbf x = \lambda \mathbf x$
- and $B = M^{-1} A \, M$
- then $A \mathbf x = A I \mathbf x = A M M^{-1} \mathbf x = \lambda \mathbf x$
- now let's multiply by $M^{-1}$ on the left:
- $\underbrace{M^{-1} A \, M}_{B} \, M^{-1} \mathbf x = M^{-1} \lambda \mathbf x$
- $B M^{-1} \mathbf x = M^{-1} \lambda \mathbf x$
- Let $\mathbf x^*$ be $M^{-1} \mathbf x$, so we have
- $B \mathbf x^* = \lambda \mathbf x^*$
- so matrices $A$ and $B$ share the same eigenvalue $\lambda$
- but not the eigenvector! $\mathbf x \ne \mathbf x^*$

For diagonalization

- $A = S^{-1} \Lambda \, S$ eigenvalues stay the same, but eigenvectors become unit vectors

What if for some $i \ne j$, $\lambda_i = \lambda_j$?

- there might be not enough eigenvectors to span $\mathbb R^n$
- i.e. columns of $A$ are not linearly independent

There are other things that make these matrices similar.

$M$ does not change:

- eigenvalues $\lambda_i$ (as discussed earlier)
- Trace and Determinant (because $\text{tr}(A) = \sum \lambda_i$ and $\text{det}(A) = \prod \lambda_i$)
- rank, and therefore number of independent eigenvectors

$M$ **does** change

- eivenvectors
- Nullspace and left nullspace
- Row Space and Column Space
- Singular Values (see SVD): they depend on $A^T A$

Suppose $A$ has a family of similar matrices

- then the Jordan Form is the most diagonal matrix of the family