The matrix

So we found the eigenvalue by doing

But how exactly do we find the dimension of the eigenspace?

Pamela Meyer
2021-12-19
Answered

How can I find the dimension of the eigenspace?

The matrix$$A=\left[\begin{array}{cc}9& -1\\ 1& 7\end{array}\right]$$ has one eigenvalue of multiplicity 2. Find this eigenvalue and the dimension of the eigenspace.

So we found the eigenvalue by doing$A-\lambda I$ to get:

$\lambda =8$

But how exactly do we find the dimension of the eigenspace?

The matrix

So we found the eigenvalue by doing

But how exactly do we find the dimension of the eigenspace?

You can still ask an expert for help

sonSnubsreose6v

Answered 2021-12-20
Author has **21** answers

The dimension of the eigenspace is given by the dimension of the nullspace of $$A-8I=(\begin{array}{cc}1& -1\\ 1& -1\end{array})$$ , which one can row reduce to $$(\begin{array}{cc}1& -1\\ 0& 0\end{array})$$ , so the dimension is 1.

Note that the number of pivots in this matrix counts the rank of A−8I. Thinking of A−8I as a linear operator from$R}^{2}\to {R}^{2$ , the dimension of the nullspace of A is given by dim$\left({R}^{2}\right)$ −rank(A)=2−1=1 by the so-called rank-nullity theorem. Of course, one can be more explicit: it is straightforward to see that the nullspace of A−8I is spanned by the vector (1,1), and hence has dimension 1.

Note that the number of pivots in this matrix counts the rank of A−8I. Thinking of A−8I as a linear operator from

Wendy Boykin

Answered 2021-12-21
Author has **35** answers

By definition, an eigenvector v with eigenvalue $\lambda$ satisfies $Av=\lambda v$ , so we have $Av-\lambda v=Av-\lambda Iv=0$ , where I is the identity matrix. Thus, $(A-\lambda I)v=0$ , and v is in the nullspace of $A-\lambda I$ .

Since the eigenvalue in your example is$\lambda =8$ , to find the eigenspace related to this eigenvalue we need to find the nullspace of A−8I, which is the matrix

$$\left[\begin{array}{cc}1& -1\\ 1& -1\end{array}\right]$$

We can row-reduce it to obtain

$$\left[\begin{array}{cc}1& -1\\ 0& 0\end{array}\right]$$

This corresponds to the equation

x - y = 0.

so x=y for every eigenvector associated to the eigenvalue$\lambda =8$ . Therefore, if (x,y) is an eigenvector, then (x,y)=(x,x)=x(1,1), meaning that the eigenspace is W=[(1,1)], and its dimension is 1.

Since the eigenvalue in your example is

We can row-reduce it to obtain

This corresponds to the equation

x - y = 0.

so x=y for every eigenvector associated to the eigenvalue

asked 2021-09-13

Assume that A is row equivalent to B. Find bases for Nul A and Col A.

asked 2021-09-18

Find an explicit description of Nul A by listing vectors that span the null space.

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For the matrix A below, find a nonzero vector in Nul A and a nonzero vector in Col A.

$A=\left[\begin{array}{cccc}2& 3& 5& -9\\ -8& -9& -11& 21\\ 4& -3& -17& 27\end{array}\right]$

Find a nonzero vector in Nul A.

$A=\left[\begin{array}{c}-3\\ 2\\ 0\\ 1\end{array}\right]$

Find a nonzero vector in Nul A.

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Finding alternate transformation matrix for similarity transformation

A pair of square matrices $X$ and $Y$ are called similar if there exists a nonsingular matrix $T$ such that ${T}^{-1}XT=Y$ holds. It is known that the transformation matrix $T$ is not unique for given $X$ and $Y$. I'm just wondering whether those non-unique transformation matrices would have any relation among themselves, like having column vectors with same directions...

What I want to mean is: Given $X$ and $Y$ a pair of similar matrices, if $S$ and $T$ are two possible transformation matrices satisfying ${S}^{-1}XS={T}^{-1}XT=Y$, is there any generic (apart from scaling) relation between $T$ and $S$ (e.g., direction of column vectors)?

For a specific example, consider $X=\left[\begin{array}{cc}A& BK\\ C& 0\end{array}\right]$ and $Y=\left[\begin{array}{cc}A+{A}^{-1}BKC& -{A}^{-1}BKC{A}^{-1}B\\ KC& -KC{A}^{-1}B\end{array}\right]$. Assuming $K$ to be invertible it can be shown that $X$ and $Y$ are similar with transformation matrix $T=\left[\begin{array}{cc}I& -{A}^{-1}B\\ 0& {K}^{-1}\end{array}\right]$. Can there be any other matrix $S$ which will be independent of $K$, and would result ${S}^{-1}XS=Y$?

A pair of square matrices $X$ and $Y$ are called similar if there exists a nonsingular matrix $T$ such that ${T}^{-1}XT=Y$ holds. It is known that the transformation matrix $T$ is not unique for given $X$ and $Y$. I'm just wondering whether those non-unique transformation matrices would have any relation among themselves, like having column vectors with same directions...

What I want to mean is: Given $X$ and $Y$ a pair of similar matrices, if $S$ and $T$ are two possible transformation matrices satisfying ${S}^{-1}XS={T}^{-1}XT=Y$, is there any generic (apart from scaling) relation between $T$ and $S$ (e.g., direction of column vectors)?

For a specific example, consider $X=\left[\begin{array}{cc}A& BK\\ C& 0\end{array}\right]$ and $Y=\left[\begin{array}{cc}A+{A}^{-1}BKC& -{A}^{-1}BKC{A}^{-1}B\\ KC& -KC{A}^{-1}B\end{array}\right]$. Assuming $K$ to be invertible it can be shown that $X$ and $Y$ are similar with transformation matrix $T=\left[\begin{array}{cc}I& -{A}^{-1}B\\ 0& {K}^{-1}\end{array}\right]$. Can there be any other matrix $S$ which will be independent of $K$, and would result ${S}^{-1}XS=Y$?

asked 2022-07-06

Take, for instance, the following matrix:

$\left[\begin{array}{cc}12& 5\\ 5& -12\end{array}\right]$

How can I find its eigenvalues/eigenvectors simply by knowing its a reflection-dilation?

$\left[\begin{array}{cc}12& 5\\ 5& -12\end{array}\right]$

How can I find its eigenvalues/eigenvectors simply by knowing its a reflection-dilation?

asked 2021-01-15

Let T denote the group of all nonsingular upper triaungular entries, i.e., the matrices of the form, [a,0,b,c] where $a,b,c\in H$

$H=\{[1,0,x,1]\in T\}$ is a normal subgroup of T.

asked 2021-05-27

Find k such that the following matrix M is singular.

$M=\left[\begin{array}{ccc}-1& -1& -2\\ 0& -1& -4\\ -12+k& -2& -2\end{array}\right]$

$k=?$