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12

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 $A$.


7

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 ...


6

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.


2

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


1

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


1

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


1

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 ...


1

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 ...


1

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 & ...


1

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 ...


1

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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