Say $A^2 = 0$. Prove that for every real number $a$, the matrix $I + a A$ is invertible.
I need some help with this question.
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Avishay28
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1Duplicate, in a particular case, of the recent question (https://math.stackexchange.com/q/2261845) – Jean Marie May 02 '17 at 10:39
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See also: If $A$ is an $n \times n$ matrix such that $A^2=0$, is $A+I_{n}$ invertible? – Martin Sleziak Jan 14 '20 at 14:53
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Let's have $v\in\ker(I+aA)$
Then $(I+aA)v=v+aAv=0$
We multiply by $A$ again to get $Av+aA^2v=Av+0=0$
But then $v+aAv=v+0=0$
So $\ker(I+aA)=\{0\}$
zwim
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Kernels related to transformations, I don't need to say ker(I+a[T]B) where B is some base of R^n? – Avishay28 May 02 '17 at 11:22
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It is common to write $\ker(M)$ for a matrix instead of $\ker(f)$ where $f$ is the linear map associated to $f(\vec x)=M\vec x$. A matrix is implicitly always associated to a linear application. – zwim May 02 '17 at 12:08
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Eigenvalues of $I+aA$ are non-zero numbers. So it is invertible.
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2Eigenvalues of $aA$ are all zeros; because $A$ is nilpotent and so those of $I+aA$ are all $1$s. – Hirakjyoti Das May 02 '17 at 10:50
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let $\lambda_1, \lambda_2,....\lambda_n$ are eigenvalues of $A_{n\times n}$, then eigenvalues of $kA$ are $k\lambda_1, k\lambda_2,....k\lambda_n$. And the eigenvalues of $I+kA$ are $1+k\lambda_1, 1+k\lambda_2,....1+k\lambda_n$. – Hirakjyoti Das May 02 '17 at 11:41
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Suppose we are given a nilpotent matrix $\mathrm A \in \mathbb R^{n \times n}$ such that $\rm A^2 = O_n$. Let its Jordan decomposition be $\rm A = P J P^{-1}$. Since $\rm A^2 = O_n$, we have $\rm J^2 = O_n$, which allows us to conclude that all the eigenvalues of $\rm A$ are zero. Let $\gamma \in \mathbb R$. Hence, all the eigenvalues of $\mathrm I_n + \gamma \mathrm A$ are $1$ and, thus, $\mathrm I_n + \gamma \mathrm A$ is invertible.