Hot answers tagged eigenvectors
4
A = $\begin{bmatrix}1-a \\ b & 1-b\end{bmatrix}$
Finding the eigenvalues and eigenvectors and writing the matrix in Jordan Normal Form yields:
$\displaystyle A = \begin{bmatrix}a & 1-a \\ b & 1-b\end{bmatrix} = P.D.P^{-1} = \begin{bmatrix}1 & \frac{a-1}{b} \\ 1 & 1\end{bmatrix} \cdot \begin{bmatrix}1 & 0 \\ 0 & a-b\end{bmatrix} ...
3
In each case, find the kernel of $A-\lambda I$. For example, when $\lambda = 1$, you have:
$$A-\lambda I = \left[\begin{array}{cccc}
2 & 0 & 0 & 0 \\
4 & 0 & 0 & 0 \\
0 & 0 & 1 & 1 \\
0 & 0 & 0 & 1 \end{array}\right]$$
Consider vectors of the form $[u,v,w,x]^{\top}$. We can multiply $A-\lambda I$ on the right ...
3
The eigenvalues and eigenvectors are:
$$\lambda_1 = -25, v_1 = (-4, 3)$$
$$\lambda_2 = 25, v_2 = (3, 4)$$
You wrote the matrix incorrectly and I used that so also made an error! I corrected it below!
For the eigenvalues, we form the system:
$$[A -\lambda I]v = 0$$
So, for $\lambda = -25$, we have:
$$\begin{bmatrix}18 & 24\\24 & ...
3
Hints:
We are given the system:
$$x' = Ax = \begin{bmatrix}
3 & -5\\
5 & 3
\end{bmatrix}x$$
with IC:
$$x(0) = \begin{bmatrix} 2\\ -3 \end{bmatrix}$$
The general solution will be:
$$x_1(t) = c_2 \sin 4 t + c_1 (3 \sin 4 t + 4 \cos 4 t)$$
$$x_2(t) = c_1 \sin 4 t + c_2 (4 \cos 4 t - 3 \sin 4 t)$$
Using the IC, will yield the final solution ...
3
Given:
$\mathbf A = \begin{bmatrix}~~3&~~2\\-9&-6\end{bmatrix}$
Part a: You should get an Eigensystem as follows:
$\lambda_1 = -3, v_1 = (-1, 3)$
$\lambda_2 = 0, v_2 = (-2, 3)$
Lets add the details for how we arrived at this first eigenvector.
$A - \lambda_1 I = 0$, yields: $\begin{bmatrix}~~6&~~2\\-9&-3\end{bmatrix}.v_1 = ...
3
We are given the matrix:
$$\begin{bmatrix}2 & 1 & 1\\1 & 2 & 1\\-2 & -2 & -1\\\end{bmatrix}$$
We want to find the characteristic polynomial and eigenvalues by solving
$$|A -\lambda I| = 0 \rightarrow -\lambda^3+3 \lambda^2-3 \lambda+1 = -(\lambda-1)^3 = 0$$
This yields a single eigenvalue, $\lambda = 1$, with an algebraic ...
2
If two matrices $ \begin{pmatrix} 3 & 2 \\ -2 & 1 \end{pmatrix}$ and $\begin{pmatrix} 1-a & -a \\ a & 1 \end{pmatrix}
$ have common eigen vector the we will have :(suppose $(x_1,x_2)$ be that eigen vector):
$\begin{pmatrix} 3 & 2 \\ -2 & 1 \end{pmatrix}\begin{pmatrix} x_1 \\ x_2 \end{pmatrix}=\lambda_1 I \begin{pmatrix} x_1 \\ x_2 ...
2
There is no promise of uniqueness for eigenvectors, (at least, not without a lot of constraints decided upon ahead of time.)
If $v$ is an eigenvector for $\lambda$, then so is $\alpha v$ for every $\alpha$ in your field $F$. It is not alarming to have different eigenvectors for a single eigenvalue.
It is also not alarming to have linearly independent ...
2
Hint: if you have
$$
Ax=\lambda x
$$
then what do you get when you add $x$ to both sides?
$$
(I+A)x = (1+\lambda)x
$$
Now do a bit of matrix algebra to get $(I+A)^{-1}$ on the right and $1+\lambda$ on the left.
(Alternative way to think of it - if $(I+A)^{-1}x = \Lambda x$, then how can you rearrange this to get $Ax$ on its own?)
2
We have the following:
$v_1 = (1, 2, 0, 0), \lambda_1 = 3$
$v_2 = (0, 0, -1, 0), \lambda_2 = 2$
$v_3 = (0, 0, 0, 0), \lambda_3 = 2 \rightarrow v_3 = {0, 0, 0, -1}$
$v_4 = (0, 1, 0, 0), \lambda_4 = 1$
Lets take an example for $\lambda_2 = 2$. We solve for:
$$|A -\lambda I|v = \begin{pmatrix}
3 - \lambda & 0 & 0 & 0 \\4 & 1- ...
2
For any square matrix with one value on the diagonal and another value everywhere else, a consistent pattern of (orthogonal) eigenvectors for the $n$ by $n$ case can be read from the columns of
$$
\left( \begin{array}{rrrrrrrrrr}
1 & -1 & -1 & -1 & -1 & -1 & -1 & -1 & -1 & -1 \\
1 & 1 ...
2
You could calculate an orthonormal basis of eigenvectors as a complex matrix. You can leave the real eigenvectors corresponding to real eigenvalues as they are.
The non-real eiegnvalues occur in complex conjugate pairs. If $\lambda$ and $\overline{\lambda}$ are one such pair, and $\lambda = re^{i\theta},$ then you can replace the diagonal submatrix ...
2
If you want results to compare against, you should get for $A = \begin{bmatrix}1 &-1 &2\\2 &-2 &4\\0 &1 &1\end{bmatrix}$, the following characteristic polynomial:
$$5 \lambda - \lambda^3 = 0$$
This CP yields the following eigenvectors and eigenvalues:
$\displaystyle \lambda_1 = -\sqrt{5}, v_1 = \left(\frac{1}{2} (-1-\sqrt{5}), ...
2
For the matrix:
$$A=\begin{bmatrix}2&1&1\\1&2&1\\1&1&2\end{bmatrix}$$
The CP is given by:
$$|A - \lambda I| = 0$$
For the CP, we get:
$$-\lambda^3+6 \lambda^2-9 \lambda+4 = -(\lambda-4) (\lambda-1)^2 = 0$$
This leads to three eigenvalues as $\lambda_1 = 4, \lambda_2 = 1, \lambda_3 = 1$.
We have a repeated eigenvalue.
To find ...
1
Calling your matrix $A$, the eigenvalue $\lambda$ and the eigenvector $y$, you use the fact that $(A-\lambda I)y=0$ So $$\begin{bmatrix}1-\sqrt 5 &-1 &2\\2 &-2-\sqrt 5 &4\\0 &1 &1-\sqrt 5\end{bmatrix}\begin{bmatrix}y_1\\y_2\\y_3 \end{bmatrix}=\begin{bmatrix}0\\0\\0 \end{bmatrix}$$ and read off three equations in three unknowns, just ...
1
The hint Michael Grant gave allows you to "split" the transformation apart into two block diagonal pieces. If you find an eigenvector for the upper left hand block, and an eigenvector for the lower right hand block, then by padding them with zeros you can make eigenvectors for the entire matrix.
So, concentrating on the lower left hand block (whose ...
1
You said that you want to find $N(A-\lambda_iI)$ so we can do it directly:
$N= \left( \begin{array}{ccc}
3-2 & 0 & 0 & 0 \\
4 & 1-2 &0 & 0 \\
0 & 0 & 2-2 & 1\\
0 &0&0&2 \end{array} \right) $
simplyfing it we get the following equalations for given $ \left( \begin{array}{ccc}
x\\
y\\
z\\
w\end{array} \right) $ ...
1
This is not quite an interpretation - rather a hint on possible generalizations.
One can look at vectors $\vec{v}=(v_x,v_y,v_z)$ in $\mathbb{R}^3$ as $2\times2$ traceless hermitian matrices $$\mathbf{v}=\left(\begin{array}{cc}v_z & v_x-iv_y \\ v_x+iv_y & -v_z\end{array}\right),$$
which form Lie algebra $su(2)$. Note that $\vec{u}\cdot ...
1
Computing the eigenvalues comes down to finding the roots of $\lambda^2 -(a+d)\lambda + (ad-bc) = 0$. That part you know already.
So if the eigenvalues are $\lambda_1$ and $\lambda_2$, then assume $c\neq 0$ and then the claim is that the eigenvectors are $v_i = (\lambda_i-d,c)$. Then
$$Av = (a\lambda_i-ad + bc, c\lambda_i - cd + cd) = (a\lambda_i - ...
1
Since $l_k^{-\frac{1}{2}}$ is not defined for $l_k=0$ I will assume $l_k\neq 0$.
Let $a_k\neq 0$ such that $YXa_k = l_k a_k$, then $$XY \left(l_k^{-\frac{1}{2}}Xa_k\right) = Xl_k^{-\frac{1}{2}}(YXa_k) = Xl_k^{-\frac{1}{2}}(l_k a_k) = l_k\left(l_k^{-\frac{1}{2}}Xa_k\right).$$
Notice that the factor $l_k^{-\frac{1}{2}}$ is actually superfluous as any nonzero ...
1
suppose that $v$ be such that $(\lambda I-A(\alpha))v=0$ then we will have $\lambda v= A_0 v+\alpha A_1 v $ $ \forall \alpha \in R$ so
$ A_0 v+2 A_1 v =\lambda v$
$ A_0 v+ A_1 v =\lambda v $
and so
$A_1 v=0$ and $A_0 v=\lambda v$
it will be correct if there exists a vector $v$
such that $v$ is an eigenvector of $A_0$ associated with the ...
1
Consider $$
A(\alpha)=\begin{pmatrix}1&\alpha\\0&-1\end{pmatrix}
=\begin{pmatrix}1&0\\0&-1\end{pmatrix}+\alpha\begin{pmatrix}0&1\\0&0\end{pmatrix}.
$$
Clearly $\lambda=-1$ is an eigenvalue for all $\alpha$, but the corresponding eigenvector is $\binom\alpha{-2}$ up to a scalar, and therefore cannot be chosen to be independent of ...
1
For the first matrix, your work is all correct.
We need to find a generalized eigenvector.
One approach to this (did you learn why in class), is to setup:
$$[A - \lambda I]v_2 = v_1$$
I am going to write the eigenvector with the signs swapped for the first one.
We have $v_1 = (-1, 1)$, so we would get:
$$[A -\lambda I]v_2 = v_1 \rightarrow ...
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