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How to prove that $$ A=\begin{bmatrix} 1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1 \end{bmatrix} $$

satisfies the equation $A^2 -4A-5I=0$?

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    $\begingroup$ If nothing else, just compute the expression.. $\endgroup$
    – Git Gud
    Apr 20, 2014 at 18:49
  • $\begingroup$ In such easy example I would simple calculate that out. $\endgroup$
    – quapka
    Apr 20, 2014 at 18:49
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    $\begingroup$ I know that but i wonder how ease as is.Thnx very much $\endgroup$ Apr 20, 2014 at 18:50
  • $\begingroup$ @Amzoti yes written correctly. $\endgroup$ Apr 20, 2014 at 18:52
  • $\begingroup$ What's $5I$?${}$ $\endgroup$
    – Shahar
    Apr 20, 2014 at 18:55

4 Answers 4

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$$A^2 = \left( \begin{array}{ccc} 9 & 8 & 8 \\ 8 & 9 & 8 \\ 8 & 8 & 9 \\ \end{array} \right)$$

$$-4A = \left( \begin{array}{ccc} -4 & -8 & -8 \\ -8 & -4 & -8 \\ -8 & -8 & -4 \\ \end{array} \right)$$

$$A^2 - 4A = \left( \begin{array}{ccc} 5 & 0 & 0 \\ 0 & 5 & 0 \\ 0 & 0 & 5 \\ \end{array} \right)$$

This is just as easy as any other way for this problem.

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  • $\begingroup$ Thnx i got what you are talking about . $\endgroup$ Apr 20, 2014 at 19:03
  • $\begingroup$ Nice explication! $\endgroup$
    – amWhy
    Apr 22, 2014 at 11:19
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$A=2J-I$, where $J$ is the all-one matrix with $J^2=3J$. Therefoer $$ A^2-4A-5I=(2J-I)^2-4(2J-I)-5I=4J^2-12J=0. $$

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  • $\begingroup$ Excellent idea thnx. $\endgroup$ Apr 20, 2014 at 19:36
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If you would like to sped up the calculation (supposing you are doing it on your own) you can use this approach.

Let's take the equation $$ A^2 - 4A - 5I = 0.$$ And now forget about the matrices. What we have is simple quadratic equation in $x$ written as $$ x^2 - 4x - 5 = 0,$$ now just find the roots $$ (x-5)(x+1) = 0.$$ Let's get back to that matrix $$(A-5I)(A+I) = 0.$$ Simple calculations give $$ \begin{pmatrix} -4 & 2 &2\\ 2 & -4 &2\\ 2 & 2 &-4 \end{pmatrix}\cdot \begin{pmatrix} 2 & 2 &2\\ 2 & 2 &2\\ 2 & 2 &2 \end{pmatrix} = 0. $$

And now it is very simple since you divide by 2 $$ \begin{pmatrix} -4 & 2 &2\\ 2 & -4 &2\\ 2 & 2 &-4 \end{pmatrix}\cdot \begin{pmatrix} 1 & 1 &1\\ 1 & 1 &1\\ 1 & 1 &1 \end{pmatrix} = 0. $$ Multiplying those matrices is just adding all the elements in each row of the first matrix. So it is $0$ everywhere. Such tricks might come in handy in more complicated examples.

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  • $\begingroup$ Excellent idea thnx. $\endgroup$ Apr 20, 2014 at 19:36
  • $\begingroup$ I am glad I could be of some help. $\endgroup$
    – quapka
    Apr 20, 2014 at 19:37
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Another way to do this is to note that $x^2 - 4x - 5 = (x-5)(x+1)$ so if we wanted to calculate this matrix $A^2 - 4A + 5I$ we can just calculate $(A-5I)(A+I)$.

This might not be much easier, especially just for a quadratic, but it does reduce the number of times you have to calculate a product of matrices, and the number of times you have to add them up.

The other ways given are perfectly good too, I just happen to prefer solving this sort of problem this way.

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