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$\require{cancel} \newcommand{\Ket}[1]{\left|{#1}\right\rangle} \newcommand{\Bra}[1]{\left\langle{#1}\right|} \newcommand{\Braket}[1]{\left\langle{#1}\right\rangle} \newcommand{\Rsr}[1]{\frac{1}{\sqrt{#1}}} \newcommand{\RSR}[1]{1/\sqrt{#1}} \newcommand{\Verti}{\rvert} \newcommand{\HAT}[1]{\hat{\,#1~}} \DeclareMathOperator{\Tr}{Tr}$
First created in February 2019
Given two matrices $A$ and $B$, if there is an invertible matrix $S$ such that $\boxed{A=SBS^{-1}}$, we say that $A$ and $B$ are similar.
It follows that $AS=SB$, which means that $S$ transforming $B$ gives the same result as $A$ transforming $S$.
??Note: Does $AS=SB$ lead to $A=SBS^{-1}$? How about $S$ not being invertible but still have $AS=SB$? Or is it that $AS=SB$ would require $S$ to be invertible? Does $A$ need to be invertible too?
Basic laws:
If $A\sim B$, $A=SBS^{-1},$ $B=S^{-1}AS=(S^{-1})A(S^{-1})^{-1}$, then $B\sim A$ (commutative law).
If $A\sim B$ and $B\sim C$, $AS_1=S_1B$, $BS_2=S_2C$, $A(S_1S_2)=S_1BS_2=S_1S_2C=(S_1S_2)C$, then $A\sim C$ (transitive law).
Because $A=IAI^{-1}$, $A\sim A$ (reflexive law).
In real numbers, $as=sb$ means $a=b$. We cannot say the same for matrices as their multiplication is not commutative, but $AS=SB$ hints that $A$ and $B$ while not equal would share some invariants.
e.g. For similar matrices,
Their determinants are equal (product of eigenvalues, see Eigenvalues).
Their characteristic polynomials are the same.
Their eigenvalues are the same (with different eigenspaces $\mathbf{v}_A=S\mathbf{v}_B$).
They are both invertable or both not.
They are both diagonalisable or both not.
The above are inter-related and can prove each other.
From the definition of similarity, $A=SBS^{-1}$.
$\det(A)=\det(SBS^{-1}) =\det(S)\det(B)\det(S^{-1}) =\det(S)\det(B){1\over\det(S)} =\det(B) .$
$\boxed{\text{Similar matrices have the same determinant}.}$
The characteristic polynomial of $A$:
$\det(\lambda I-A) =\det\left(S(\lambda I-B)S^{-1}\right) =\det(S)\det(\lambda I-B)\det(S^{-1}) =\det(\lambda I-B).$
$\boxed{\text{Similar matrices have the same characteristic polynomial}.}$
Similar matrices have the same characteristic polynomial.
$\det(\lambda I-A) =\det(\lambda SIS^{-1}-SBS^{-1}) =\det(S(\lambda I-B)S^{-1}) =\det(S)\det(\lambda I-B)\det(S^{-1}) =\det(\lambda I-B).$
Eigenvalues are the zeros of characteristic polynomial. i.e. $\boxed{\text{Similar matrices have the same set of eigenvalues}.}$
If $\mathbf{v}_B$ is an eigenvector of $B, B\mathbf{v}_B=\lambda\mathbf{v}_B,~~ AS\mathbf{v}_B=SB\mathbf{v}_B=\lambda S\mathbf{v}_B,~~ A(S\mathbf{v}_B)=\lambda(S\mathbf{v}_B).$
i.e. $\boxed{\text{If }\mathbf{v}_B\text{ is an eigenvector of }B,~S\mathbf{v}_B\text{ is an eigenvector of }A.~}$
For a full rank $B$ with $\{\mathbf{v}_B\}$ being its eigenbasis, $SBS^{-1}$ convert it into a new matrix $A$ with $\{S\mathbf{v}_B\}$ as eigenbasis.
Let us consider $S$ being a transformation transforming basis vectors of $B$ into those of $A$. (The basis does not necessarily be an eigenbasis.)
Now $S^{-1}AS\Ket\psi$ means to transform $\Ket\psi$ from basis of $B$ into basis of $A$, apply $A$, then reverse-transform the result back to the basis of $B$, effectively $B\Ket\psi$.
Example: To implement $cZ$ using $cX$, you may transform the basis of $Z~~\{\Ket0,\Ket1\}$ by $H$ into the basis of $X~~\{\Ket+,\Ket-\}$, perform the $cX$ operation, then transform back to $\{\Ket0,\Ket1\}$ by $H^{-1}=H$. In short, $cZ=H~cX~H.$
Let $P=\big[\mathbf{v}_1\mathbf{v}_2\ldots\mathbf{v}_n\big]$, where $\mathbf{v}_r$ are eigenvectors of $A$. Also let $D$ be a diagonal matrix with the eigenvalues of $A$ on its diagonal, in the same order as $\mathbf{v}_r$.
As $A\mathbf{v}_r=\lambda_r\mathbf{v}_r,~~ A\big[\mathbf{v}_1\mathbf{v}_2\ldots\mathbf{v}_n\big] =\big[A\mathbf{v}_1~A\mathbf{v}_2\ldots A\mathbf{v}_n\big] =\big[(\lambda_1\mathbf{v}_1)(\lambda_2\mathbf{v}_2)\ldots(\lambda_n\mathbf{v}_n)\big] =\big[\mathbf{v}_1\mathbf{v}_2\ldots\mathbf{v}_n\big]D .$ We have $AP=PD$ so $A$ and $D$ are similar.
If a matrix $A$ is similar to a diagonal matrix $D$, then $A$ is said to be diagnoisable.
Let $D= \begin{bmatrix} d_1 & 0 & \ldots & 0 \\ 0 & d_2 & \ldots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \ldots & d_n \end{bmatrix}.$
Two observations:
All $d_r$ ($r=1,2,\ldots,n$) are eigenvalues of $D$, as $\det(d_r I-D)=\prod_{k=1}^n(d_r-d_k)=0$.
The product of its eigenvalues on the diagonal is the determinant of $D$. i.e. $\det(D)=\prod_{k=1}^nd_r.$
The diagonal matrix $D$ that $A$ is similar to has its determinant being the product of all its eigenvalues. As $A$ is sharing the same eigenvalues and the same determinant with $D$, it immediately follows that the $\boxed{\text{product of eigenvalues of a matrix is its determinant}.}$
By rearranging the order of $\mathbf{v}_r$, we will have $D'$, with the same set of eigenvalues as $D$ but in different order.
Let $P'$ be the eigenvector matrix rearranged. By the same process, we will find $AP'=P'D'$ so $A$ and $D'$ are similar.
By the transitive law, $D$ and $D'$ are similar. $D'=P'^{-1}AP'=(P'^{-1}P)~D~(P^{-1}P').$
This means all invertible matrices that are similar to each other forms a "conjugacy class". In fact, each invertible matrices belongs to precisely one conjugacy class of similarity.
There are several related concepts about linear maps with domain of $n$ dimension (e.g. a square matrix $A$ of $n\times n$). In the following, vector $\mathbf{v}$ is in the $n$-D domain.
Rank - the dimension of space the $n$ column vectors of $A$ span.
In order words, it is the dimension of the image of $A$ (the space of all possible $A\mathbf v$).
When rank is $n$, we call it "full rank".
Nullity - the dimension of space that $A$ gives a zero projection.
In order words, it is the dimension of the "kernel" of $A$ (the space of $\mathbf v$ such that $A\mathbf v=0$).
A "full rank" $A$ has a only the zero vector in its kernel.
The sum of Rank and Nullity is $n$, the dimension of the domain - Rank-Nullity Theorem.
To visualise the above, imaging a real projector matrix of 3x3, mapping any 3D vectors onto a 2D plane. The image is a 2D plane, so rank is 2. The kernel is the ray at the origin perpendicular to the image plane, so the nullity is 1. 2+1=3, which is the dimension of the domain.
The number of non-zero elements in a diagonal matrix is obviously its rank.
For $A$ if there is $B$ such that $AB=BA=I_n$, $A$ is said to be invertible, or non-singular or nondegenerate.
?? The followings are necessary and sufficient conditions of each other:
$A$ is invertible.
$\det(A)\ne 0$.
$A$ is full rank.
$(1)\Rightarrow(2)$: $\det(AB)=\det(A)\det(B)=\det(I)=1$, so not $(2)$ means not $(1)$. $(2)\Rightarrow(1)$ can resort to well-known process to obtain $A^{-1}$ like Cayley–Hamilton method.
?? A full rank $A$ is similar to a diagonal matrix $D$ with $n$ non-zero eigenvalues.
A matrix being degenerate means that it is not invertible, while eigenvalues being degenerate means they are equal.
| Diagonal | Matrix Degenerate | Eigenvalues Degenerate |
|---|---|---|
| 1, 2, 3 | No | No |
| 1, 2, 0 | Yes | No |
| 1, 2, 2 | No | Yes |
| 2, 2, 0 | Yes | Yes |
WORKING
Degeneracy is related to rank and nullity but not the same concept. A diagonal matrix with 1, 2, 2 on the diagonal is full rank but
?? Not all matrices are diagonalisable. But if a matrix has $n$ eigenvectors (including identical ones), it is diagonalisable. In fact, as long as $P$ is invertible (i.e. there are $n$ mutually independent eigenvectors, $A$ does not need to be invertible.
?? If there are identical eigenvalues, $D$ is not invertible, neither is $A$, but it may still have $n$ mutually independent eigenvectors, so $P^{-1}$ exists and therefore is diagonalisable (i.e. similar to $D$).
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