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Given: $$A \in M_{nxn} (\mathbb C), \; A \neq \lambda I, \; A^2 + 2A = 3I$$

Now we define: $$B = A^2 + A - 6I$$

The question:

Is $B$ inversable?

Now, what I did is this:

$A^2 + 2A = 3I \rightarrow \lambda^2v + 2\lambda v = 3v \rightarrow \lambda_1 = 1, \lambda_2 = -3$

Is what I suggested correct? I know that if so, I just do the same to B and calculate the determinant.

Gottfried Helms
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TheNotMe
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    Question 1: What are the eigenvalues of $A$? Question 2: What are the eigenvalues of $A^2+A-6I ?$ – Start wearing purple May 02 '13 at 10:42
  • Question 1: Those are the ones that I calculated, IF they exist, no? Question 2: I just apply B to v and see what I get? – TheNotMe May 02 '13 at 10:44
  • Note that the factorization $A^2+2A-3I=(A-I)(A+3I)$ implied by your solution of the corresponding quadratic equation can also be used to obtain the equation $(A-I)(A^2+A-6I)=(A-2I)(A^2+2A-3I)$ used in robjohn's answer. – joriki May 05 '13 at 16:08

5 Answers5

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Note that $$ \begin{align} (A-I)B &=(A-I)(A^2+A-6I)\\ &=(A-2I)(A^2+2A-3I)\\[4pt] &=0 \end{align} $$ Thus, unless $A=I$, and therefore, $B=-4I$, $(A-I)B=0$ implies that $B$ is not invertible.

Clarification: Suppose that $B^{-1}$ exists, then $$ \begin{align} A-I &=(A-I)BB^{-1}\\ &=0B^{-1}\\[6pt] &=0 \end{align} $$ Thus, if $B^{-1}$ exists, then $A=I$. This is the contapositive of "if $A\ne I$, then $B$ is not invertible".

A bit of explanation

I used the Euclid-Wallis Algorithm to try and write $(A^2+A-6I)x+(A^2+2A-3I)y=I$ to compute an inverse for $A^2+A-6I\bmod A^2+2A-3I$: $$ \begin{array}{c|c} &&1&-A+I\\\hline 1&0&1&A-I\\ 0&1&-1&-A+2I\\ A^2+A-6I&A^2+2A-3I&-A-3I&0 \end{array} $$ Unfortunately, this showed that $A+3I$ was the GCD of $A^2+A-6I$ and $A^2+2A-3I$. However, this did show that $$ (A-I)(A^2+A-6I)-(A-2I)(A^2+2A-3I)=0 $$ which was used in the answer above.

Community
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robjohn
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  • why is it $=0$? – TheNotMe May 02 '13 at 13:56
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    @TheNotMe: because we are given $A^2+2A=3I$ – robjohn May 02 '13 at 14:23
  • If $A\neq I$ then $(A-I)B = 0$ implies that $B=0$ thus its' determinant $=0$ so it is not invertible, and we know that $A\neq I$ so $B$ is for sure $=0$? – TheNotMe May 02 '13 at 14:46
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    @TheNotMe: I am not appealing to determinants. If $B^{-1}$ exists, then $$\begin{align}A-I&=(A-I)BB^{-1}\&=0B^{-1}\&=0\end{align}$$ Thus, if $B^{-1}$ exists, then $A=I$. – robjohn May 02 '13 at 14:50
  • Note that $(A-I)(A^2+A-6I)=(A-2I)(A^2+2A-3I)$ can also be found using the factorization $A^2+2A-3I=(A-I)(A+3I)$ implicit in the OP's solution of the corresponding quadratic equation. – joriki May 05 '13 at 16:06
  • @joriki: thanks for that; it's nice to have several approaches. I was not trying to imply that what I had done was the only way; I was simply explaining how I came across the answer I gave. – robjohn May 05 '13 at 16:21
  • Sure :-) I just wanted to point that out because it implies that the OP had made more progress towards a solution than it might have seemed. – joriki May 05 '13 at 17:04
  • @joriki: Ah, nice. I hadn't yet seen your comment to the OP. Thanks. – robjohn May 05 '13 at 17:10
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$B=A^{2}+A-6I$

$B=(A+3I)(A-2I)$

$det(B)=det(A+3I)det(A-2I)$

now attention :

$A^{2}+2A=3I$

$A^2+2A-3I=0$ $\to$ $det(A+3I)det(A-I)=0$

$A^2+2A-3I-4A+4I=-4A+4I$

$(A-I)^2=-4(A-I)$ $\to$ $if det(A-I)=-4$ then B willnot be inversable

in other word if -3 or 2 be eigen value of A then B will not be inversable

Somaye
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  • indeed this question tell us if 1 not be eigen value of A then -3 will be eigen value and B will not be inversable – Somaye May 02 '13 at 12:27
  • I did not understand why $(A-I)^2 = -4(A-I)$ means that $ifdet(A-I) = -4$ then $B$ will not be inversable. – TheNotMe May 02 '13 at 14:31
  • if det(A-I) not be equal t0 0 then det(A+3I) must be equal 0 (because det(A+3I)det(A-I)=0) and so B will not be inversable – Somaye May 02 '13 at 18:39
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Hint: Using Jordan normal form for $A$.

Ma Ming
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  • We have not learned this and I do not think this is the method I should go through! Although I looked it up in wikipedia and it seems very interesting. Thanks, though! – TheNotMe May 02 '13 at 10:45
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We have $$B=-3I-A$$ and we can see easily that the spectrum of $B$ is $$\mathrm{sp}(B)=\{-3-\lambda_1,-3-\lambda_2\}=\{-4,0\}$$ so $B$ is not invertible.

Added Since $A^2+2A-3I=(A+3I)(A-I)=0$ and since $A\not= \lambda I$ then the minimal polynomial of is $x^2+2x-3$ which is a product of distinct linear factors over $\mathbb{R}$ then $A$ is diagonalisable and $-3$ and $1$ are it's eigenvalues hence there's $P\in\mathrm{GL}_n(\mathbb{R})$ s.t. $A=PDP^{-1}$ where $D$ is a diagonal matrix and then we have $$B=-3I-A=-3PP^{-1}-PDP^{-1}=P(-3I-D)P^{-1}$$ so we can deduce the spectrum of $B$ from the spectrum of $A$ as explained above.

  • Can you please explain how did you calculate the spectrum? (sorry for the stupid question) – TheNotMe May 02 '13 at 10:50
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    How do you know that $0$ is indeed present in the spectrum of $B$? – Start wearing purple May 02 '13 at 10:55
  • Ok, how do you know that $-3$ is indeed present in the spectrum of $A$? – Start wearing purple May 02 '13 at 11:00
  • Okay so we know the spectrum is B. And again, excuse me for ANOTHER stupid question (new to diagonalization), why does ${-4,0}$ as a spectrum of a matrix $B$ means that $B$ is not inversable? – TheNotMe May 02 '13 at 11:03
  • @Sami Ben Romdhane: Sorry, but all eigenvalues of $A$ equal to 1 doesn't imply $A$ is equal to the identity matrix. – Start wearing purple May 02 '13 at 11:07
  • @TheNotMe: The only problem is that "$0$ is in the spectrum" is not yet proven. – Start wearing purple May 02 '13 at 11:09
  • Wait wait. Let us ask this question: Is what I did in the beginning really correct? I mean, what if A has no real eigenvalues? Why am I allowed to calculate eigenvalues if I know not for sure that they exist, as real numbers? – TheNotMe May 02 '13 at 11:10
  • @SamiBenRomdhane: What about $\left(\begin{array}{cc} 1 & 1 \ 0 & 1\end{array}\right)$? Who are its eigenvalues? Do you know that there exist non-diagonalisable matrices? – Start wearing purple May 02 '13 at 11:13
  • that all its eigenvalues are equal to 1 and it's not the identity matrix, obviously. – Start wearing purple May 02 '13 at 11:17
  • Can I just say, that regarding the $D$ matrix we found for $B$, that $B$ has a line of zeros, thus, its' determinant is equal $0$, thus, it is non-invertible? – TheNotMe May 02 '13 at 11:22
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    @SamiBenRomdhane: You've been assuming $A$ is diagonalizable, and other folks haven't. In fact, because the minimal polynomial has distinct roots, $A$ must be diagonalizable. – Ted Shifrin May 02 '13 at 11:24
  • So, I assume that what you did is correct? – TheNotMe May 02 '13 at 11:24
  • @TedShifrin In our case $A$ is diagonalisable as you explained by the minimal polynomial but the method remains the same even $A$ is not diagonalisable since it's at least triangularisable in $\mathbb{C}$ –  May 02 '13 at 11:30
  • @Sami Ben Romdhane: I suggest you delete your answer or complete your reasoning. In the present form it is incomplete and simply wrong. – Start wearing purple May 02 '13 at 11:32
  • Once again: what is written presently is wrong, call it hint or whatever you like. – Start wearing purple May 02 '13 at 11:41
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$A$ is a root of the polynomial $x^2 + 2x - 3=(x-1)(x+3).$

Since $A$ is not of the form $\lambda I,$ this is the minimal polynomial of $A.$ Consequently $-3$ is an eigenvalue of $A.$

$$\color{red}{B=A^2 + A - 6I=A^2+2A-3I-A-3I=-A-3I}$$

$-3$ is an eigenvalue of $A\implies\det(A+3I)=0\implies\det(-B)=0\implies\det B=0.$

Sugata Adhya
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