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* 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 | ||
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* 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 $M$ | + | |
− | + | ||
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* 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 $i \ne j$, $\lambda_i = \lambda_j$? | ||
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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
Invariant subspace
Invariant column space
Suppose $A$ has all its eigenvalues
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 $B$?
Let's check it:
For diagonalization
What if for some $i \ne j$, $\lambda_i = \lambda_j$?
There are other things that make these matrices similar.
$M$ does not change:
$M$ does change
Suppose $A$ has a family of similar matrices