# Showing that matrix is invertible using eigenvalues

Let $A$ be matrix from the vector space of square $N \times N$ matrices.

With the inital information: $A^2-4A=4I$.

How does one show that $A+I$ is invertible?

(I need please a solution that involves eigenvalues)

Thank you

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I guess the homework tag would be appropriate... –  Fabian Mar 27 '11 at 21:40
@Qiaochu: I did not try alot. Actually I don't know how eigenvalues relate to that. –  user6163 Mar 27 '11 at 21:50
@Fabian- I'm an acadmy returner after a while, I do that for refreshing my memory. When I know how to approach to an ex, and think that I can handle it by myself I'm adding the tag. –  user6163 Mar 27 '11 at 21:52

Note that a square matrix $B$ is invertible if and only if $\lambda=0$ is not an eigenvalue of $B$.

Thus, \begin{align*} A+I\text{ is invertible }&\Longleftrightarrow \lambda=0\text{ is not an eigenvalue of }A+I\\ &\Longleftrightarrow \text{there is no nonzero solution to }(A+I)x=0\\ &\Longleftrightarrow \text{there is no nonzero solution to }Ax +x = 0\\ &\Longleftrightarrow \text{there is no nonzero solution to }Ax = -x\\ &\Longleftrightarrow \lambda = -1\text{ is not an eigenvalue of }A. \end{align*}

The "initial information" tells you that $A$ satisfies $t^2-4t-4$. So the minimal polynomial of $A$ divides $(t-2-2\sqrt{2})(t-2+2\sqrt{2})$.

What does that tell you about the eigenvalues of $A$?

Added. More generally, $\lambda$ is an eigenvalue of $A$ if and only if $\lambda+\mu$ is an eigenvalue of $A+\mu I$. Because $Ax=\lambda x$ if and only if $(A-\mu I)x = (\lambda-\mu)x$. So $A+\mu I$ is invertible if and only if $-\mu$ is not an eigenvalue of $A$.

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Arturo I think you have a mistake in the polynomial that $A$ satisfies. Although that doesn't alter your answer =) –  Adrián Barquero Mar 27 '11 at 21:47
My answer might have confused you; the equation is $t^2-4t-4=0$ -- I mis-placed a minus sign, so the eigenvalues are different (the point is still the same though). Sorry. –  InterestedGuest Mar 27 '11 at 21:50
@Adrian: Thank you; @Interested: No, I misplaced the minus sign all on my own, didn't need your help to screw that up. (-: –  Arturo Magidin Mar 27 '11 at 21:51

You have $A^2-4A-4I=0$ for any $\vec{x}$. So, if $\lambda$ is an eigenvalue of $A$, then for some corresponding eigenvector, you can rewrite the equation as ${\lambda}^{2}\vec{x}-4\lambda\vec{x}-4\vec{x}=0$. Factoring $\vec{x}$ out, you get $({\lambda}^{2}-4\lambda-4)\vec{x}=0$. Since you know that an eigenvector cannot be the zero vector, you know that ${\lambda}^{2}-4\lambda-4$ has to equal zero. From here on, find roots to the equation -- you can use Viete's formulas; it factors as $(\lambda-2(1-\sqrt(2))(\lambda-2(1+\sqrt(2))=0$, giving you two eigenvalues of $A$. So, since $0$ is not an eigenvalue of $A$, you know that $A$ is invertible (and so is $A^{2}$). From the original equation, you also know that $A^{2}=4A+4I$, so $A+I=\frac{A^{2}}{4}$. Since $A^{2}$ is invertible, and dividing a matrix by its scalar does not affect its invertibility (determinant can't become 0, preservation of dimensions etc. remains the same), you have that $A+I$ is equal to an invertible matrix. Hence, $A+I$ is invertible.

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doesn't the fact that 2 is the only eigenvalue of $A$ already tell you that $A+I$ is invertible? –  Fabian Mar 27 '11 at 21:44
@Fabian: good point; I took it the long way. –  InterestedGuest Mar 27 '11 at 21:47
@Fabian: The eigenvalues of $A$ are $2 \pm 2\sqrt{2}$, not 2. –  Hans Lundmark Mar 28 '11 at 7:15
@Hans: I caused the confusion by misplacing a minus sign in the original equation (and getting 2 as the only eigenvalue). –  InterestedGuest Mar 28 '11 at 14:05

Hint: What does the initial information tell you about the eigenvalues of $A$?

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