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I got stuck solving this problem:

Let $T:\mathbb{R}^3\to \mathbb{R}^3$ be the linear transformation defined by the matrix A in the standard basis of $\mathbb{R}^3$, $E=\{e_1,e_2,e_3\}$ $$A=\begin{bmatrix} 3 & 1 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 2 \end{bmatrix}$$

and let $W=\ker(T-3I)$ be a subspace of $\mathbb{R}^3$, Show that there is no $T$ invariant subspace $U$ of $\mathbb{R}^3$ that satisfies $\mathbb{R}^3=W \oplus U$

Note: when solving this problem I encountered this problem:
Direct sum of subspaces of the three dimensional space

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  • $\begingroup$ Have you determined ker(T-3I) yet? $\endgroup$ – Bernd Jan 9 '14 at 10:56
  • $\begingroup$ Yes: $ker(T-3I) = span\{e_1\}$ (where $e_1=(1,0,0)$) $\endgroup$ – MathNerd Jan 9 '14 at 10:58
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    $\begingroup$ You were right with this $\endgroup$ – Bernd Jan 9 '14 at 11:25
  • $\begingroup$ human sometimes make mistake $\endgroup$ – dato datuashvili Jan 9 '14 at 11:26
  • $\begingroup$ but wait it should be invariant subspace right?because kernel span vector and eigenvector span vector is the same $\endgroup$ – dato datuashvili Jan 9 '14 at 11:29
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kernel of $ (T-3*I)$ is any vector of the form $(0,x,0)$ which is spanned by $A=(0,1,0)$ is not it?

Edited: we we have like this

A=[3 1 0;0 3 0;0 0 2]

A =

 3     1     0
 0     3     0
 0     0     2

and we have done A-3*eye(3)

ans =

 0     1     0
 0     0     0
 0     0    -1

where

eye(3)

ans =

 1     0     0
 0     1     0
 0     0     1

clearly this matrix

A-3*eye(3)

ans =

 0     1     0
 0     0     0
 0     0    -1

has solution

$0*x_1+1*x_2+0*x_3=0$

$0*x_1+0*x_2+0*x_3=0$

$0*x_1+0*x_2+(-1)*x_3=0$

we get

$x_3=0$

$x_2=0$

$x_1=k$

yes exactly it is spanned by $e_1=(1,0,0)$

we have

Let v be an eigenvector of T, i.e. T v = λv. Then W = span {v} is T invariant. As a consequence of the fundamental theorem of algebra, every linear operator on a complex finite-dimensional vector space with dimension at least 2 has an eigenvector. Therefore every such linear operator has a non-trivial invariant subspace. The fact that the complex numbers are algebraically closed is required here. Comparing with the previous example, one can see that the invariant subspaces of a linear transformation are dependent upon the underlying scalar field of V.

[V D]=eig(A)

V =

1.0000   -1.0000         0
     0    0.0000         0
     0         0    1.0000

D =

 3     0     0
 0     3     0
 0     0     2

see please what is output of eigenvalue decomposition

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  • $\begingroup$ No. I think JohnSaita determined the kernel correctly $\endgroup$ – Bernd Jan 9 '14 at 11:03
  • $\begingroup$ directly how? i did not understand $\endgroup$ – dato datuashvili Jan 9 '14 at 11:04
  • $\begingroup$ I corrected it - but I just saw, that the y - coordinate can also be arbitrary, so it's not correct. hang on $\endgroup$ – Bernd Jan 9 '14 at 11:05
  • $\begingroup$ Sorry. The kernel you determined is correct $\endgroup$ – Bernd Jan 9 '14 at 11:09
  • $\begingroup$ sorry @Bernd see please update $\endgroup$ – dato datuashvili Jan 9 '14 at 11:19

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