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Since $\lambda$ is an eigenvalue of $A^2$, we know that $$\det (A^2 - \lambda I) = 0$$ From here we conclude that $$\det (A^2 - \lambda I) = \det((A - \sqrt{\lambda}I)(A + \sqrt{\lambda}I)) = \det(A - \sqrt{\lambda}I) \times\det ( A + \sqrt{\lambda}I)= 0$$ Hence $\sqrt{\lambda}$ or $-\sqrt{\lambda}$ is an eigenvalue of $A$.


First note that: $$A^2 - \lambda I = (A-\sqrt\lambda I)(A+\sqrt\lambda I)$$ Let $v$ be an eigenvector of $A^2$ with eigenvalue $\lambda$. We can use $v$ to find an explicit eigenvector of $A$ with eigenvalue that is either $\sqrt\lambda$ or $-\sqrt\lambda$. Since $(A^2-\lambda I)v = 0$, we must have either $(A+\sqrt\lambda I)v = 0$, in which case $v$ is ...


Well, the set of all eigenvectors of an eigenvalue forms a subspace. I.e. if u,v $\in Eig(A,\lambda)$ then $a_1 u$+$a_2v\in Eig(A,\lambda)$: For $Av=\lambda v$ and $Au=\lambda u$ we have: $A(a_1u+a_2v)=a_1Au+a_2Av=a_1\lambda u+a_2\lambda v=\lambda(a_1u+a_2v)$ Note: $0$ is not an eigenvector. So the set above is only a space if we add $0$ to the space.


Here's a positive definite counterexample: $$\begin{bmatrix}2&1&0\\1&2&0\\0&0&1\end{bmatrix}.$$


you should treat it as a normal polynominal when you are trying to find the roots (which are eigenvalues). $$\det(A-\lambda \cdot I) = (\lambda-4)(\lambda+2)^2$$


$$\det(A-\lambda I) = (1-\lambda)(-2-\lambda)(1-\lambda) + 3(2+\lambda)3=\\ =(1-2\lambda + \lambda^2)(-2-\lambda) + 9(2+\lambda) = \\ =(-2-\lambda + 4\lambda + 2\lambda^2 - 2\lambda^2 - \lambda^3) + 18 + 9\lambda = \\ = -\lambda^3 + 3\lambda - 2 + 18 + 9\lambda = -\lambda^3 + 12\lambda + 16,$$ not what you got...


Remember that an eigenvector defines a subspace of the domain of the linear transformation. That subspace has many bases --- in fact, each nonzero multiple of the eigenvector is also a basis for the eigenspace. So both $[2,5]$ and $[-2,-5]$ are bases, and either can be thought of as "representing" the one-dimensional eigenspace. The choice of $[-2,-5]$ is ...


Any multiple of an eigenvector is still an eigenvector for the same eigenvalue, even if this multiple is negative. So if $(2,5)^T$ is an eigenvector then so are $(-2,-5)^T$, $(10,25)^T$, $(1,5/2)^T$, $(-6,-15)^T$, ... If the machine marking the answers is clever enough (for example it's using MapleTA) it should accept any of these. As to why it has chosen ...


It follows from the Jordan decomposition that if $v_{\lambda^2}$ is an eigenvector of $A^2$ with eigenvalue $\lambda^2$, then it is either an eigenvector of $A$ corresponding to the eigenvalue $\lambda$ resp. $-\lambda$, or a combination $v_{\lambda}+v_{-\lambda}$. To see this, just square the Jordan block $$ \begin{pmatrix} \lambda & 1 & 0 & ...


If the geometric multiplicity of $0$ is two which means that the dimension of the eigenspace of $0$ is $2$ then there's two linearly independent eigenvectors associated to $0$ and then the given matrix would be similar to $$\begin{pmatrix}0&0&0\\0&0&1\\0&0&0\end{pmatrix}$$ and this is a contradiction. Think to the rank of the two ...


The notation $A^H$ means the hermitian (or conjugate) transpose of $A$. You want to show that, for any vector $v\in N(A)$ and any vector $w\in C(A^H)$, the (standard) inner product $$ v^Hw=0 $$ The definition of $C(A^H)$ says that $w=A^Hu$ for some $u$; then $$ v^Hw=v^HA^Hu=(Av)^Hu=0 $$ because, by assumption, $v\in N(A)$.

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