Construct a 2x2 matrix with real eigenvalues that is not diagonalizable
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I've been banging my head against the table with this one for a while. I know that for it not to be diagonalizable that the columns can't be linearly independent but can't quite seem to come up with one.
linear-algebra eigenvalues-eigenvectors diagonalization
asked Apr 3 '17 at 3:01
uRockNinja
111
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up vote
2
down vote
favorite
I've been banging my head against the table with this one for a while. I know that for it not to be diagonalizable that the columns can't be linearly independent but can't quite seem to come up with one.
linear-algebra eigenvalues-eigenvectors diagonalization
asked Apr 3 '17 at 3:01
uRockNinja
111
3
You might want to consider a $2times 2$ Jordan block.
– thanasissdr
Apr 3 '17 at 3:05
1
What you think you know might not be correct.
– Brian Borchers
Apr 3 '17 at 3:08
For it not to be diagonalizable all you need is to have an eigenvalue with multiplicity greater than the dimension of the eigenspace. (i.e not enough eigenvectors). This is not related to independence
– B.A
Apr 3 '17 at 3:23
Also complex matrices can always be put on a form with either 1s or 0s off the main diagonal and eigenvalues on the diagonal. For real matrices you can be sure to not have to have more than at most blocks of size 2x2 along the diagonal.
– mathreadler
Apr 3 '17 at 9:11
add a comment |
up vote
2
down vote
favorite
up vote
2
down vote
favorite
I've been banging my head against the table with this one for a while. I know that for it not to be diagonalizable that the columns can't be linearly independent but can't quite seem to come up with one.
linear-algebra eigenvalues-eigenvectors diagonalization
asked Apr 3 '17 at 3:01
uRockNinja
111
I've been banging my head against the table with this one for a while. I know that for it not to be diagonalizable that the columns can't be linearly independent but can't quite seem to come up with one.
linear-algebra eigenvalues-eigenvectors diagonalization
linear-algebra eigenvalues-eigenvectors diagonalization
asked Apr 3 '17 at 3:01
uRockNinja
111
asked Apr 3 '17 at 3:01
uRockNinja
111
asked Apr 3 '17 at 3:01
uRockNinja
111
asked Apr 3 '17 at 3:01
uRockNinja
111
asked Apr 3 '17 at 3:01
uRockNinja
111
111
3
You might want to consider a $2times 2$ Jordan block.
– thanasissdr
Apr 3 '17 at 3:05
1
What you think you know might not be correct.
– Brian Borchers
Apr 3 '17 at 3:08
For it not to be diagonalizable all you need is to have an eigenvalue with multiplicity greater than the dimension of the eigenspace. (i.e not enough eigenvectors). This is not related to independence
– B.A
Apr 3 '17 at 3:23
Also complex matrices can always be put on a form with either 1s or 0s off the main diagonal and eigenvalues on the diagonal. For real matrices you can be sure to not have to have more than at most blocks of size 2x2 along the diagonal.
– mathreadler
Apr 3 '17 at 9:11
add a comment |
3
You might want to consider a $2times 2$ Jordan block.
– thanasissdr
Apr 3 '17 at 3:05
1
What you think you know might not be correct.
– Brian Borchers
Apr 3 '17 at 3:08
For it not to be diagonalizable all you need is to have an eigenvalue with multiplicity greater than the dimension of the eigenspace. (i.e not enough eigenvectors). This is not related to independence
– B.A
Apr 3 '17 at 3:23
Also complex matrices can always be put on a form with either 1s or 0s off the main diagonal and eigenvalues on the diagonal. For real matrices you can be sure to not have to have more than at most blocks of size 2x2 along the diagonal.
– mathreadler
Apr 3 '17 at 9:11
3
3
You might want to consider a $2times 2$ Jordan block.
– thanasissdr
Apr 3 '17 at 3:05
You might want to consider a $2times 2$ Jordan block.
– thanasissdr
Apr 3 '17 at 3:05
1
1
What you think you know might not be correct.
– Brian Borchers
Apr 3 '17 at 3:08
What you think you know might not be correct.
– Brian Borchers
Apr 3 '17 at 3:08
For it not to be diagonalizable all you need is to have an eigenvalue with multiplicity greater than the dimension of the eigenspace. (i.e not enough eigenvectors). This is not related to independence
– B.A
Apr 3 '17 at 3:23
For it not to be diagonalizable all you need is to have an eigenvalue with multiplicity greater than the dimension of the eigenspace. (i.e not enough eigenvectors). This is not related to independence
– B.A
Apr 3 '17 at 3:23
Also complex matrices can always be put on a form with either 1s or 0s off the main diagonal and eigenvalues on the diagonal. For real matrices you can be sure to not have to have more than at most blocks of size 2x2 along the diagonal.
– mathreadler
Apr 3 '17 at 9:11
Also complex matrices can always be put on a form with either 1s or 0s off the main diagonal and eigenvalues on the diagonal. For real matrices you can be sure to not have to have more than at most blocks of size 2x2 along the diagonal.
– mathreadler
Apr 3 '17 at 9:11
add a comment |
1 Answer
1
active
oldest
votes
up vote
0
down vote
Try
$$
begin{bmatrix}1&1\0&1end{bmatrix}
$$
with repeat eigenvalue $1$ already in its Jordan form.
For this to be diagonalizable, we need two distinct eigenvectors for the eigenvalue $1$, i.e. an eigenbasis that spans the domain. But
$$
ker(begin{bmatrix}0&1\0&0end{bmatrix})=begin{bmatrix}1\0end{bmatrix}
$$
with $dim(ker (begin{bmatrix}1&1\0&1end{bmatrix})=1<2$.
answered Apr 3 '17 at 3:06
qbert
21.5k32457
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
Try
$$
begin{bmatrix}1&1\0&1end{bmatrix}
$$
with repeat eigenvalue $1$ already in its Jordan form.
For this to be diagonalizable, we need two distinct eigenvectors for the eigenvalue $1$, i.e. an eigenbasis that spans the domain. But
$$
ker(begin{bmatrix}0&1\0&0end{bmatrix})=begin{bmatrix}1\0end{bmatrix}
$$
with $dim(ker (begin{bmatrix}1&1\0&1end{bmatrix})=1<2$.
answered Apr 3 '17 at 3:06
qbert
21.5k32457
add a comment |
up vote
0
down vote
Try
$$
begin{bmatrix}1&1\0&1end{bmatrix}
$$
with repeat eigenvalue $1$ already in its Jordan form.
For this to be diagonalizable, we need two distinct eigenvectors for the eigenvalue $1$, i.e. an eigenbasis that spans the domain. But
$$
ker(begin{bmatrix}0&1\0&0end{bmatrix})=begin{bmatrix}1\0end{bmatrix}
$$
with $dim(ker (begin{bmatrix}1&1\0&1end{bmatrix})=1<2$.
answered Apr 3 '17 at 3:06
qbert
21.5k32457
add a comment |
up vote
0
down vote
up vote
0
down vote
Try
$$
begin{bmatrix}1&1\0&1end{bmatrix}
$$
with repeat eigenvalue $1$ already in its Jordan form.
For this to be diagonalizable, we need two distinct eigenvectors for the eigenvalue $1$, i.e. an eigenbasis that spans the domain. But
$$
ker(begin{bmatrix}0&1\0&0end{bmatrix})=begin{bmatrix}1\0end{bmatrix}
$$
with $dim(ker (begin{bmatrix}1&1\0&1end{bmatrix})=1<2$.
answered Apr 3 '17 at 3:06
qbert
21.5k32457
Try
$$
begin{bmatrix}1&1\0&1end{bmatrix}
$$
with repeat eigenvalue $1$ already in its Jordan form.
For this to be diagonalizable, we need two distinct eigenvectors for the eigenvalue $1$, i.e. an eigenbasis that spans the domain. But
$$
ker(begin{bmatrix}0&1\0&0end{bmatrix})=begin{bmatrix}1\0end{bmatrix}
$$
with $dim(ker (begin{bmatrix}1&1\0&1end{bmatrix})=1<2$.
answered Apr 3 '17 at 3:06
qbert
21.5k32457
edited Apr 3 '17 at 3:30
answered Apr 3 '17 at 3:06
qbert
21.5k32457
answered Apr 3 '17 at 3:06
qbert
21.5k32457
answered Apr 3 '17 at 3:06
qbert
21.5k32457
21.5k32457
add a comment |
add a comment |
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3
You might want to consider a $2times 2$ Jordan block.
– thanasissdr
Apr 3 '17 at 3:05
1
What you think you know might not be correct.
– Brian Borchers
Apr 3 '17 at 3:08
For it not to be diagonalizable all you need is to have an eigenvalue with multiplicity greater than the dimension of the eigenspace. (i.e not enough eigenvectors). This is not related to independence
– B.A
Apr 3 '17 at 3:23
Also complex matrices can always be put on a form with either 1s or 0s off the main diagonal and eigenvalues on the diagonal. For real matrices you can be sure to not have to have more than at most blocks of size 2x2 along the diagonal.
– mathreadler
Apr 3 '17 at 9:11