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## Hot answers tagged eigenvalues-eigenvectors

4

An $n \times n$ matrix has a characteristic polynomial of order $n$. If the matrice's elements are from an algebraically closed field, then that means it must have at least one root. Then we can factor out a root and the remaining factor must also have at least one root et.c. Therefore there must be $n$ roots. The characteristic polynomial has eigenvalues as ...

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$P$ will diagonalize $A$ if the columns of $P$ are eigenvectors of $A$. It looks like the 3rd column of $P$ is an eigenvector for the eigenvalue 7, so then you just need to see if the other three columns span the eigenspace for the eigenvalue 1.

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Recall that the squares of the singular values are the eigenvalues of $A^*A$. So $$B = \begin{pmatrix} A \\ b \end{pmatrix} \implies B^*B = A^*A+ b^*b ,$$ and the eigenvalues increase (though perhaps not strictly) by the min-max principle.

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Yes, it is invertible, because it is positive definite: Let $B=D+L-L^T$ where $D$ is the diagonal matrix with positive entries and $L$ is the lower part of the skew symmetric matrix $A$ , and $-L^T$ is the upper part. Then you have $$(Bx,x)=((D+L-L^T)x,x)=(Dx,x)+(Lx,x)-(L^Tx,x)=(Dx,x)+(L^Tx,x)-(L^Tx,x)=(Dx,x)>0$$ for each nonzero vector $x$. We used ...

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I will assume $\mathcal{P}(\mathbb{R})$ is the set $\mathbb{R}[x]$ of polynomials with real coefficients. Then if $p$ is an eigenvector of $T$, we have $T p = p - p' = \lambda p$. By comparing the leading coefficients, we have that $\lambda =1$, so that $p'=0$, and $p=c$ for some constant $c$. Thus there is only one eigenvalue, $\lambda=1$, with ...

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Affirmative if you are taking matrices over the complex numbers, although not necessarily distinct. Over the real numbers the answer is sometimes a natural number $k$ between $0$ and $n$

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Hint: Let $x$ be an eigenvector of $B$. Let $A$ be a matrix with zeros in every column except for the $j$th column, and take the $j$th column to be the vector $x$.

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Matrices will commute when operating on any linear combination of shared eigenvectors if $A\vec v_i=a_i \vec v_i$ and if $B\vec v_i=b_i \vec v_i$ then an arbitrary linear combination of shared eigenvectors is $\sum_i \alpha_i \vec v_i$ $$AB(\sum_i \alpha_i \vec v_i)=(\sum_i a_i b_i\alpha_i \vec v_i)=BA(\sum_i \alpha_i \vec v_i)$$ if the vectors $v_i$ ...

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The solution to a linear ODE system $x'=Ax,x(0)=x_0$ (where $A$ is a constant square matrix) is given by $x(t)=e^{At}x_0$. This should not be particularly surprising in view of the 1D case, but it also is not very useful by itself, because it is not obvious how to compute $e^{At}$, which is defined as $\sum_{k=0}^\infty \frac{t^k A^k}{k!}$. The most ...

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The first is a classical Jordan form of $2\times2$, and the second diagonal, both with $0$ as the eigenvalue. For $A$ eigenvalue multiplicity is $2$. For $B$ multiplicity $1$. Consider also, both the characteristic and minimal polynomials of them: $$\chi_A(x)=\mu_A(x)=x^2$$ and $$\chi_B(x)=x^2\quad,\quad \mu_B(x)=x$$

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Not sure this is exactly what you want, but it's a linear transformation without explicitly specifying a matrix, or action on a basis, and also finding (real) eigenvalues. Given a vector $v$ in ${\bf R}^3$, define the linear transformation $T_v:{\bf R}^3\to{\bf R}^3$ by $T_v(w)=v\times w$ (the cross product). Now suppose $T_v(w)=\lambda w$. We know ...

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Another example: consider the linear transformation on the vector space of infinitely differentiable functions given by $T(f)=f'$. You get the eigenvalues (and eigenvectors) by solving the differential equation $${dy\over dx}=\lambda y$$ Every real number $\lambda$ is an eigenvalue, with corresponding eigenvector $e^{\lambda x}$.

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There are only two roots since a polynomial of degree $n$ has exactly $n$ roots (in $\mathbb{C}$). Here, $n=2$, so there are only $2$ roots. The statement in your second question is not correct. For example, consider the identity matrix of dimension $n$. Clearly the identity matrix is diagonalizable (as it is diagonal), but it has characteristic polynomial ...

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Actually the characteristic polynomial can't have a degree higher than 3 in this case, but we can say for sure that the characteristic polynomial must divide that larger polynomial so it must share the factors in some combination. So we still need to pick some configuration from $1$ and $-2$. Trace is sum of eigenvalues. What combinations of sums of $1$ and ...

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Think about (simple) eigenspaces. The eigenvalue tells you the scaling factor for how that matrix acts on the corresponding eigenspace. (There are some details about generalized eigenvalues that aren't super relevant here) So in the eigenvalue 0 case, the eigenvector tells you what subspace is getting collapsed to the origin.

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You do not have to calculate $A+A^2+A^3$. Suppose that you know how to diagonalize the matrix $A$, i.e., you can find an invertible matrix $P$ and a diagonal matrix $D$ such that $A=PDP^{-1}$. (I will leave the computations to you. You can have a look at other posts tagged diagonalization or on Wikipedia. I guess you can find there something to get you ...

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Hint: Let $A$ be the $3\times 3$ matrix. If $\mathbf{v_1}=\begin{bmatrix} x_1\\x_2\\x_3\end{bmatrix}$ is the eigenvector that corresponds to the eigenvalue $\lambda_1 =1$, in order to find the eigenspace $V_{\lambda_1}$ we may solve the system: $$A\cdot \begin{bmatrix} x_1\\x_2\\x_3\end{bmatrix} = 1\cdot \begin{bmatrix} x_1\\x_2\\x_3\end{bmatrix},$$ with ...

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The eigenspace associated with the eigenvalue $\lambda$ will be the set of all solutions to the equation $$(A-\lambda I)x=0$$ One of your eigenvalues is $3$, let's look at that one. What we need to do here is solve $$(A-3I)x=0$$ So first off, what's $A-3I$? It's \pmatrix{4 & 0 & 1 \\ -2 & 1 & 0 \\ -2 & 0 & 1} - \pmatrix{3 & ...

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The elements of your numerical eigenvector all have the same sign, so there is a representative of the eigenspace which has all nonnegative entries (e.g. the one taken by multiplying all the entries of your numerical eigenvector by $-1$). Of course the representative of probabilistic interest is the one with all nonnegative entries and which has a sum of ...

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Thanks for your response A.G. I can't seem to respond to your comment as I'm an unregistered user (a guest). Yes, there is a lot of dependence. There is a matrix U that block diagonalises A; if $$U = \frac{1}{\sqrt{2}} \begin{pmatrix} I & -J \\ J & I \end{pmatrix},$$ then U^TAU = \begin{pmatrix} 0 ...

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