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Proof for Diagonal Matrices from Page 2 of 7: Let $A \in M_{n}(C)$ be diagonal, to wit, $A _{ii}=\lambda_{i}$.
Then $ p_{A}(t) = \det(tI-A)= \det \begin{bmatrix} t - \lambda_1 & ~ & ~ \\ ~ & \ddots & ~ \\ ~ & ~ & t - \lambda_n \\ \end{bmatrix} =\prod_{i=1}^{n}(t-\lambda_{i}) \quad (♦)$
and $p_A(A)= \prod_{i=1}^{n}(A- \color{forestgreen}{ \lambda_iI } ) $ , a product of diagonal matrices.

$1.$ How does $p_A(A)= \prod_{i=1}^{n}(A-\lambda_{i}I) $ ? $(♦)$ contains $\lambda_i$ and NOT $\color{forestgreen}{ \lambda_iI }$ ?
What legitimates this? $t$ is a variable but A is a matrix, so they can't be equal?

Does the proof repeat this technique for the last line of this proof, denoted with $\color{ orangered }{ ( \yen ) }$?

As in the previous examples (on the linked PDF in the first sentence), since $A$ is diagonal, $$ p_{A}(A) \mathop{=}^{\color{ red }{\clubsuit} } \begin{bmatrix} p_A(\lambda_1) & ~ & ~ \\ ~ & \ddots & ~ \\ ~ & ~ & p_A(\lambda_n) \\ \end{bmatrix} = \begin{bmatrix} \prod_{i=1}^{n}(\lambda_1 -\lambda_{i}) & ~ & ~ \\ ~ & \ddots & ~ \\ ~ & ~ & \prod_{i=1}^{n}(\lambda_n -\lambda_{i}) \\ \end{bmatrix} = \text{ 0 matrix },$$ where $\prod_{i=1}^{n}(\lambda_n -\lambda_{i}) = ...(\lambda_n -\lambda_{n})= 0 $, and the same holds for all the other diagonal entries.

$2.$ How does $p_{A}(A)$ equal that diagonal matrix, as denoted with $\color{ red }{ ( \clubsuit )} $ ?

Proof for Diagonalisable Matrices: Similar matrices have the same eigenvalues (and thus characteristic polynomials), so suppose for similar matrices A and $B$ (now $A$ may NOT be diagonal): $ p_{A}(z)=p_{B}(z)=\displaystyle \sum_{i=0}^{n}c_{i}z^{i} \implies p_{A}(A)=\sum_{i=0}^{n}c_{i}A^{i} \quad \color{ orangered }{ ( \yen ) } $ (I omit the rest of the proof.)

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  • $\begingroup$ The statement of Cayley Hamilton (for a general matrix) says precisely that when you substitute $A$ for the variable in the characteristic polynomial of $A$, and evaluate the resulting matrix, you will get the zero matrix. No, indeed, $A$ is not the same as the variable $t$, but the resulting product of matrices does make sense. If $A$ is diagonal matrix, then for each eigenvalue $\lambda$ of $A,$ we can consider the matrix $A- \lambda I$. If we take the product of these matrices, as $\lambda$ runs through all eigenvalues of $A$, we get the zero matrix. $\endgroup$ May 17, 2014 at 20:17
  • $\begingroup$ In general, if $D={\rm diag}(a_1,\ldots,a_n)$; $P(A)={\rm diag}(P(a_1),\ldots,P(a_n))$. $\endgroup$
    – Pedro
    May 17, 2014 at 21:43
  • $\begingroup$ @PandaBear Better? $\endgroup$
    – user53259
    May 22, 2014 at 17:15

2 Answers 2

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I believe what is happening is that the two "functions" $p_A(t)$ and $p_A(A)$ are defined by a product. $\prod_{i=1}^n(t-\lambda_i)$,$\prod_{i=1}^n(A-\lambda_iI)$,

In $1$ Dimension, $p_A(t)=\det(tI-A)$, i.e. in this case the product and the determinant are the same.

Clearly this is not the same in higher dimensions.

Now $p_A(A)$ is itself a diagonal matrix (since it is the product of diagonal matrices).

This is because $p_A(A)=\prod_{i=1}^n(A-\lambda_iI)$, for each $i$, $(A-\lambda_iI)$ is a diagonal matrix (because both $A$ and $I$ are).

Thus $p_A(A)=\prod_{i=1}^n(A-\lambda_iI)=(A-\lambda_1I)...(A-\lambda_nI)$, is a product of diagonal matrices.

We see that the $j$-th diagonal entry is: $\prod_{i=1}^n(A_{jj}-\lambda_i)=\prod_{i=1}^n(\lambda_j-\lambda_i)=p_A(\lambda_j)$.

This because:

$p_A(A)=\prod_{i=1}^n(A-\lambda_iI)=(A-\lambda_1I)...(A-\lambda_nI)$

$=\left[\begin{array}{ccc} a_{11}-\lambda_1 & 0 & ... & 0\\ 0 & a_{22}-\lambda_1& ...& 0 \\ ...& ...& ...& ...\\ 0 & ... & ...& a_{nn}-\lambda_1 \end{array} \right]...\left[\begin{array}{ccc} a_{11}-\lambda_n & 0 & ... & 0\\ 0 & a_{22}-\lambda_n& ...& 0 \\ ...& ...& ...& ...\\ 0 & ... & ...& a_{nn}-\lambda_n \end{array} \right]$

$=\left[\begin{array}{ccc} (a_{11}-\lambda_1)...(a_{11}-\lambda_n) & 0 & ... & 0\\ 0 & (a_{22}-\lambda_1)...(a_{22}-\lambda_n)& ...& 0 \\ ...& ...& ...& ...\\ 0 & ... & ...& (a_{nn}-\lambda_1)...(a_{nn}-\lambda_n) \end{array} \right]$

$=\left[\begin{array}{ccc} \prod_{j=1}^n(a_{11}-\lambda_j) & 0 & ... & 0\\ 0 & \prod_{j=1}^n(a_{22}-\lambda_j)& ...& 0 \\ ...& ...& ...& ...\\ 0 & ... & ...& \prod_{j=1}^n(a_{nn}-\lambda_j) \end{array} \right]$

$=\left[\begin{array}{ccc} \prod_{j=1}^n(\lambda_1-\lambda_j) & 0 & ... & 0\\ 0 & \prod_{j=1}^n(\lambda_2-\lambda_j)& ...& 0 \\ ...& ...& ...& ...\\ 0 & ... & ...& \prod_{j=1}^n(\lambda_n-\lambda_j) \end{array} \right]$

$=\left[\begin{array}{ccc} p_A(\lambda_1) & 0 & ... & 0\\ 0 & p_A(\lambda_2)& ...& 0 \\ ...& ...& ...& ...\\ 0 & ... & ...& p_A(\lambda_n) \end{array} \right]$

From this we see that each diagonal entry in the matrix correspond to what you have posted.

Please ask if anything is unclear.

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They are related in the sense that substituting $A$ for $t$ in $p_A(t)$ will give you $p_A(A)$, but the latter is an $n×n$ matrix, whereas, $p_A(t)$ is a scalar, so they only truly coincide if A is a one dimensional matrix.

The reason that substituting $A$ for $t$ in $p_A(t)$ gives us $p_A(A)$, is as follows,

Let $t=A$ (and in the formula change $\lambda_i$ to $\lambda_i I$ so that we can add, subtract and multiply with the correct dimensions), and we get:

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You are correct here, because the dimensions don't agree, we need to let $t$ be the matrix $A$ and let $\lambda_i$ be $\lambda_i I$ so really we should have:

$t\to A$, $\lambda_i\to \lambda_i I$

$p_A(t)=\prod_{i=1}^n(t-\lambda_i)\to\prod_{i=1}^n(A-\lambda_i I)=p_A(A)$.

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  • $\begingroup$ It is not quite that $p_A(A)$ is defined as a product. If $p(x)=a_0+a_1x+\dots+a_nx^n$, you define $p(A)=a_0I+a_1A+\dots+a_nA^n$, and then you need to verify that if $p=qr$ (as polynomials), then indeed $p(A)=q(A)r(A)$ (as matrices). In particular, this gives us that for any two polynomials $q,r$, the matrices $q(A)$ and $r(A)$ commute. $\endgroup$ May 26, 2014 at 17:02
  • $\begingroup$ @LePressentiment, hopefully this edit should help :) $\endgroup$
    – Ellya
    May 26, 2014 at 17:38
  • $\begingroup$ +1. Thank you. I see 2 now, thanks to you. But for 1, so is $p_A(A)$ NOT related to $p_A (t)$? Must I memorise $p_A(A)$ as the definition? Is that what Andres Caicedo is saying? $\endgroup$
    – user53259
    May 27, 2014 at 11:48
  • $\begingroup$ They are related in the sense that substituting $A$ for $t$ in $p_A(t)$ will give you $p_A(A)$, but the latter is an $n\times n$ matrix, whereas, $p_A(t)$ is a scalar, so they only truly coincide if $A$ is a one dimensional matrix. $\endgroup$
    – Ellya
    May 27, 2014 at 12:18
  • $\begingroup$ @LePressentiment, I have put in an edit, I hope it helps :) $\endgroup$
    – Ellya
    May 28, 2014 at 9:15
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In answer to 1) you can substitute $t$ for $A$ because we can evaluate $p(x)$ in an algebra over $\mathbb{C}$ (in this case). And yes it does repeat in the last bit of the proof.

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  • $\begingroup$ Were you able to answer my question 2? Would you also please enlarge on your answer? $\endgroup$
    – user53259
    May 22, 2014 at 17:14

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