# Tag Info

1

If $A$ is an $m\times n$ matrix, then also $\Sigma$ is $m\times n$, so $U$ must be $m\times m$ and $V$ is $n\times n$. So if you want that $U$ and $V$ have the same shape, then $A$ must be square. You're right in thinking that having $U=V$ implies the (square) matrix $A$ is normal: actually it must be Hermitian. If the SVD is $A=U\Sigma U^\dagger$, then $A^... 1 You have done everything correct except for the small mistake in the last step. While normalising the vector, you have multiplied by the norm instead of dividing it and hence you are getting different values. For the first column, instead of$\frac{\sqrt{14}}{2}$the term should be$\frac{2}{\sqrt{14}}$. And similarly for the other column. That should fix ... 0 Orthogonality is defined by the specific inner product being used. The expression of that inner product in terms of coordinates is naturally dependent on the basis. However, one can always find an orthonormal basis (that’s what the Gram-Schmidt process does) and relative to that basis, the inner product of two vectors is equal to the dot product of their ... 2 To diagonalize, first notice that $$A\begin{bmatrix}1\\1\\1\end{bmatrix}=\begin{bmatrix}3a\\3a\\3a\end{bmatrix}$$ so $$u:=\begin{bmatrix}1\\1\\1\end{bmatrix}$$ is an eigenvector of the$3a$eigenvalue. To find eigenvectors$v,w$of the$0$eigenvalue, simply reduce$A$to its RREF and determine a basis for the kernel. Then we can diagonalize$A$as$A=P^{-1}...

0

It is a decomposition similar to the Cholesky called the $LDL$ decomposition import numpy as np from scipy.linalg import ldl A = np.array([[3,-1,0],[-1,2,-1],[0,-1,3]]) A array([[ 3, -1, 0], [-1, 2, -1], [ 0, -1, 3]]) lu, d, perm = ldl(A, lower =0) lu array([[ 1. , -0.6 , 0. ], [ 0. , 1. , -0....

3

Your matrix $X$ is not orthogonal: the columns do not have norm $1$. Divide each column by its norm, and all will be fine then. Not being orthogonal means that $X^{-1}\neq X^T$.

1

For $$P^{−1}AP= Λ$$ your matrix $P$ is the matrix whose columns are the eigenvectors of $A$ and Λ is the diagonal matrix of eigenvalues. Note that we need linearly independent eigenvectors in order for the matrix $P$ to be invertible.

0

There are two fundamental steps in the proof: the first relies on $A$ being real and symmetric, thus admitting a diagonalization, $Q^T A Q = \Lambda$, through a change of coordinates $z=Q^Tx$ with $Q$ being an orthogonal matrix (satisfying $Q^{-1} = Q^T$); being $Q$ an orthogonal matrix, it preserves distance in the transformation ( isometry), that is $$\|... 1 You have correctly calculated two eigenvectors of H. In order to produce an orthonormal basis of eigenvectors (to get the corresponding unitary matrix), you want your eigenvectors v^{(1)}, v^{(2)} to satisfy:$$\left \langle v^{(1)}, v^{(2)} \right \rangle := v^{(1)}_1 \overline{v^{(2)}_1} + v^{(1)}_2 \overline{v^{(2)}_2} = 0 \ \ \ \text{ and}  \...

1

Notice that $A$ satisfies the equation $x^3-x^2-x+1=0$. Upon factoring we see that the eigenvalues are $1$ and $-1$, where $1$ is of multiplicity $2$. So if $A$ is diagonalizable, we can write $A$ as $A=PDP^{-1}$, where the diagonal matrix $D$ is given by \begin{pmatrix} 1 & 0 & 0\\ 0 & 1 &0 \\ 0 & 0 & -1 \end{pmatrix} So, if we ...

2

Suppose $A$ is a diagonalizable, and write $A=PDP^{-1}$. Then the equation becomes $PD^3P^{-1}=PD^2P^{-1}+PDP^{-1}-I$, which when we multiply by $P^{-1}$ on the left and $P$ on the right, we see is equivalent to $D^3=D^2+D-I$. Thus it suffices to show the diagonal case. Let $D$ have diagonal entries $\lambda_i$ ($1\leq i\leq n$). Then for each $i$ we have ...

-1

In case anyone's looking for a detailed proof of diagonalization of circulant matrices and their eigenvalues and eigenvectors, below powerpoint has a great explanation between slides 37-45: www.cs.uoi.gr/~cnikou/Courses/Digital_Image_Processing/2010-2011/Chapter_04c_Frequency_Filtering_(Circulant_Matrices).ppt

2

Hint If $\lambda$ is an eigenvalue, then $$\lambda^3-\lambda^2-\lambda+1=0 \Rightarrow (\lambda-1)^2(\lambda+1)=0 \Rightarrow \lambda=\pm 1$$ What is then $D^2$? Therefore $$A^2=PD^2P^{-1}=??$$

-1

Hint: consider the 1 dimensional case, which is immediate. Then argue that this case is sufficient by considering the fact that $A$ is diagonalisable.

0

Yes, your answer is correct. The matrix $A$ is similar to a diagonal matrix $D$ which has no other entries in the main diagonal than $1$, $-1$, $2$, and $-2$. Therefore, its minimal polynomial (which is the minimal polynomial of $D$) is the one that you mentioned.

0

If $A$ were diagonalizable then your diagonal matrix would be identity matrix $I$ and you would have $A=PIP^{-1} = I$. So if $A$ is not identity matrix and has $n$ eigenvalues $=1$ then it is not diagonalizable.

0

The Eigenvectors are such that $$Am_i=m_i d_i$$ where $d_i$ is the associated Eigenvalue. Write this relation for all Eigenvectors and pack in a matrix, to get $$AM=MD.$$ $D$ is a digonal matrix, the effect of which is to multiply every column vector by the corresponding diagonal coefficient. If the Eigenvectors are linearly independent, the matrix $M$...

2

Assume that we have $n$ distinct eigenvalues for the $n\times n$ matrix $A$ then we can formulate the eigenvalue equations for all the eigenvalue and eigenvector pairs as $$Av_i=\lambda_iv_i \text{ for } i=1,2,...,n.$$ We can assemble all these equations into one single matrix equation $$A[v_1,...,v_n]=[v_1,...,v_n]\text{diag}[\lambda_1,...,\lambda_n]$$ ...

0

Consider a complex nxn matrix A. In this case, A being normal implies the Schur Decomposition gives a diagonal matrix. This says that we can diagonalize A using orthogonal vectors. And therefore the matrix A has n linearly independent eigen vectors and hence it is semi-simple.

2

If $A$ had just one eigenvalue, it wouldn't be diagonalizable because its algebraic multiplicity would be equal to $\,n\,$ (dimension of the whole vector space), but its geometric multiplicity: $\dim E_{\lambda_1}=n-1<n$ However, since $A$ has two different eigenvalues, their algebraic multiplicity must be at least $1$. So, if $\dim\left(E_{\... 0 Note that the eigenspace for$A$corresponding to$\lambda=1$is the two-dimensional$\{s(0,0,1)+t(-1,0,1)\colon s,t\in\Bbb R\}$; in particular, there are other eigenvectors besides the ones you listed. Similarly for$B$and$\lambda=0$: the eigenspace is$\{s(-1,1,0)+t(0,0,1)\colon s,t\in\Bbb R\}$. So you have more possibilities for the eigenvectors from ... 7 Using that your matrices are skew symmetric, you get that these matrices are diagonalizable. See the section spectral theory on this Wikipedia article. 31 All those matrices are anti-symmetric and therefore they are normal matrices. And every normal matrix is diagonalizable over$\mathbb C$, by the spectral theorem. 30 The matrix$A_n$is a tridiagonal Toeplitz matrix with diagonal entries$\delta = 0$and off-diagonal entries$\tau = 1$and$\sigma = -1$. Hence, we can use the formula in this paper to show that the eigenvalues are $$\lambda_k = 2i\cos\left(\dfrac{k\pi}{n+1}\right),$$ for$k = 1,\ldots,n$, and the corresponding eigenvectors$v_1,\ldots,v_n$have entries$\$...

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