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We have the matrix $A \in R^{2\times2}$ and we know

$(A-5I_{2})^5=0_2$

Can we invert $A$?

  • 1
    Doesn't your equation imply $A=5I$ ? – 2017-02-05
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    @AhmedS.Attaalla: no, take for example $$\begin{pmatrix} 5& 1 \\ 0&5 \end{pmatrix}.$$ – 2017-02-05

4 Answers 4

4

Hint: any matrix commutes with $I$ so by binomial expansion:

$$(A-5I_{2})^5=0_2 \;\;\implies\;\; A\cdot(A^4-5\cdot5 A^3+10\cdot5^2A^2-10\cdot5^3A + 5\cdot5^4 I_2) = 5^5 I_2$$

3

$$B=A-5I→B^5=0$$

Suppose that $\lambda$ is an eigenvalue of $B$ then

$$BX=\lambda X→B^5X=\lambda^5X→\lambda=0$$

It means that all eigenvalues of $B$ is $0$.

You can suppose that you can't invert $A$ then it has $0$ as an eigenvalue, so $AY=0→BY=-5Y$ what is impossible because all eigenvalues of $B$ are zero.

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    Arnaldo thank you very much, are you sure about this solution? – 2017-02-05
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    @MaketheDif: yes, I am. – 2017-02-05
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    You are very welcome! – 2017-02-05
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    Downvote?!!! Is anything wrong? – 2017-02-05
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    Please give an explanation of why you downvoted to let the poster know whats wrong. – 2017-02-05
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    The argument is basically correct (see [my answer](http://math.stackexchange.com/a/2131369/18880) which I wrote before reading this), but it is not very well formulated here. Don't say "the eigenvector" and "the eigenvalue" when they are not unique (or not yet known to be). Also $k$ for an eigenvalue and $X,Y$ for eigenvectors are awkward names. But +1 anyway. – 2017-02-06
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    Hi @MarcvanLeeuwen, thanks for the hint. – 2017-02-06
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    What names people usualy use for eigenvector? I'm very familiar with $X$. – 2017-02-06
2

Since \begin{eqnarray} p(x)&=&(x-5)^5=\sum_{k=0}^5{5\choose k}(-5)^kx^{5-k}=x^5-25x^4+250x^3-1250x^2+3125x-3125, \end{eqnarray} we have $$ p(A)=(A-5I)^5=A^5-25A^4+250A^3-1250A^2+3125A-3125I=0. $$ It follows that $$ A^5-25A^4+250A^3-1250A^2+3125A=3125I, $$ i.e. $$ A\left[\dfrac{1}{3125}\left(A^4-25A^3+250A^2-1250A+3125I\right)\right]=I. $$ Hence $A$ is invertible and $$ A^{-1}=\dfrac{1}{3125}\left(A^4-25A^3+250A^2-1250A+3125I\right). $$

2

If $A$ were non-invertible it would have an eigenvalue$~0$. But $(A-5I)^5$ would act on a corresponding eigenvector by the scalar $(-5)^5\neq0$, which would contradict $(A-5I)^5=0$. So yes, $A$ is invertible.

The question does not ask for an inverse, just whether we can invert. But in case you are curious, here it is. Since the matrix is $2\times 2$ one already has $(A-5I)^2=0$, that is $A^2-10A+25I=0$. Then one easily checks that $A-10I$ is a matrix that multiplied by$~A$ gives$~{-}25I$, so $A^{-1}=\frac25I-\frac1{25}A$.