Solution to this problem is incorrect about a diagonalizable matrix A?
I have this matrix A
$$ begin{bmatrix}
1 & c & 0 & 0 & 0 \
0 & 1 & c & 0 & 0 \
0 & 0 & 1 & a & 0\
0 & 0 & 0 & b & c\
0 & 0 & 0 & 0 & b
end{bmatrix}
$$
(d) (4 points) Find all values of a, b, c, if any, such that A is
diagonalizable but not orthogonally diagonalizable. Justify your
answer.
Solution: By spectral theorem, we know that A is orthogonally
diagonalizable iff A is symmetric. So we know that at least one of a
or c must be nonzero so that A is not symmetric. If b !=1, we know
by part (c) that c=0 for the matrix to be diagonalizable. Therefore, a != 0.
If b = 1, then the matrix is only diagonalizable iff a = c = 0 which
is a contradiction as then A will be symmetric. Therefore, the
requirements are the b != 1, c = 0, and a != 0.
However, b must be 0. For example, you cannot have b = 2. If b = 2, then the dimension of the kernel of this matrix is only $1$, so the eigenvalue $1$ has $gemu(1) < almu(1)$, so this matrix would not be diagonalizable. Is there anything wrong with my reasoning?
linear-algebra matrices diagonalization
edited Dec 6 at 9:25
José Carlos Santos
148k22117218
asked Apr 18 at 23:13
Goldname
1,450928
add a comment |
I have this matrix A
$$ begin{bmatrix}
1 & c & 0 & 0 & 0 \
0 & 1 & c & 0 & 0 \
0 & 0 & 1 & a & 0\
0 & 0 & 0 & b & c\
0 & 0 & 0 & 0 & b
end{bmatrix}
$$
(d) (4 points) Find all values of a, b, c, if any, such that A is
diagonalizable but not orthogonally diagonalizable. Justify your
answer.
Solution: By spectral theorem, we know that A is orthogonally
diagonalizable iff A is symmetric. So we know that at least one of a
or c must be nonzero so that A is not symmetric. If b !=1, we know
by part (c) that c=0 for the matrix to be diagonalizable. Therefore, a != 0.
If b = 1, then the matrix is only diagonalizable iff a = c = 0 which
is a contradiction as then A will be symmetric. Therefore, the
requirements are the b != 1, c = 0, and a != 0.
However, b must be 0. For example, you cannot have b = 2. If b = 2, then the dimension of the kernel of this matrix is only $1$, so the eigenvalue $1$ has $gemu(1) < almu(1)$, so this matrix would not be diagonalizable. Is there anything wrong with my reasoning?
linear-algebra matrices diagonalization
edited Dec 6 at 9:25
José Carlos Santos
148k22117218
asked Apr 18 at 23:13
Goldname
1,450928
add a comment |
I have this matrix A
$$ begin{bmatrix}
1 & c & 0 & 0 & 0 \
0 & 1 & c & 0 & 0 \
0 & 0 & 1 & a & 0\
0 & 0 & 0 & b & c\
0 & 0 & 0 & 0 & b
end{bmatrix}
$$
(d) (4 points) Find all values of a, b, c, if any, such that A is
diagonalizable but not orthogonally diagonalizable. Justify your
answer.
Solution: By spectral theorem, we know that A is orthogonally
diagonalizable iff A is symmetric. So we know that at least one of a
or c must be nonzero so that A is not symmetric. If b !=1, we know
by part (c) that c=0 for the matrix to be diagonalizable. Therefore, a != 0.
If b = 1, then the matrix is only diagonalizable iff a = c = 0 which
is a contradiction as then A will be symmetric. Therefore, the
requirements are the b != 1, c = 0, and a != 0.
However, b must be 0. For example, you cannot have b = 2. If b = 2, then the dimension of the kernel of this matrix is only $1$, so the eigenvalue $1$ has $gemu(1) < almu(1)$, so this matrix would not be diagonalizable. Is there anything wrong with my reasoning?
linear-algebra matrices diagonalization
edited Dec 6 at 9:25
José Carlos Santos
148k22117218
asked Apr 18 at 23:13
Goldname
1,450928
I have this matrix A
$$ begin{bmatrix}
1 & c & 0 & 0 & 0 \
0 & 1 & c & 0 & 0 \
0 & 0 & 1 & a & 0\
0 & 0 & 0 & b & c\
0 & 0 & 0 & 0 & b
end{bmatrix}
$$
(d) (4 points) Find all values of a, b, c, if any, such that A is
diagonalizable but not orthogonally diagonalizable. Justify your
answer.
Solution: By spectral theorem, we know that A is orthogonally
diagonalizable iff A is symmetric. So we know that at least one of a
or c must be nonzero so that A is not symmetric. If b !=1, we know
by part (c) that c=0 for the matrix to be diagonalizable. Therefore, a != 0.
If b = 1, then the matrix is only diagonalizable iff a = c = 0 which
is a contradiction as then A will be symmetric. Therefore, the
requirements are the b != 1, c = 0, and a != 0.
However, b must be 0. For example, you cannot have b = 2. If b = 2, then the dimension of the kernel of this matrix is only $1$, so the eigenvalue $1$ has $gemu(1) < almu(1)$, so this matrix would not be diagonalizable. Is there anything wrong with my reasoning?
linear-algebra matrices diagonalization
linear-algebra matrices diagonalization
edited Dec 6 at 9:25
José Carlos Santos
148k22117218
asked Apr 18 at 23:13
Goldname
1,450928
edited Dec 6 at 9:25
José Carlos Santos
148k22117218
asked Apr 18 at 23:13
Goldname
1,450928
edited Dec 6 at 9:25
José Carlos Santos
148k22117218
edited Dec 6 at 9:25
José Carlos Santos
148k22117218
edited Dec 6 at 9:25
José Carlos Santos
148k22117218
148k22117218
asked Apr 18 at 23:13
Goldname
1,450928
asked Apr 18 at 23:13
Goldname
1,450928
asked Apr 18 at 23:13
Goldname
1,450928
1,450928
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
Why do you think that $b$ must be $0$? That's not true. Take, for instance this case:$$A=begin{pmatrix}1 & 0 & 0 & 0 & 0 \ 0 & 1 & 0 & 0 & 0 \ 0 & 0 & 1 & 1 & 0 \ 0 & 0 & 0 & 2 & 0 \ 0 & 0 & 0 & 0 & 2end{pmatrix}.$$It is diagonalizable: for instance, the vectors $(1,0,0,0,0)$, $(0,1,0,0,0)$, and $(0,0,1,0,0)$ are eigenvalues with eigenvalue $1$ and $(0,0,1,1,0)$ and $(0,0,0,0,1)$ are eigenvectors with eigenvalue $2$.
answered Apr 18 at 23:34
José Carlos Santos
148k22117218
Ah I made mistake. I forgot c = 0.
– Goldname
Apr 19 at 1:11
add a comment |
With the given constraints on $a$ and $c$, $det A=b^2$, so its kernel is always trivial for $bne0$. If instead by the first “this matrix” you mean $$A-bI = begin{bmatrix}1&0&0&0&0\0&1&0&0&0\0&0&1&a&0\0&0&0&0&0\0&0&0&0&0end{bmatrix},$$ its null space is quite clearly two-dimensional regardless of the value of $a$.
answered Apr 18 at 23:50
amd
29k21050
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
Why do you think that $b$ must be $0$? That's not true. Take, for instance this case:$$A=begin{pmatrix}1 & 0 & 0 & 0 & 0 \ 0 & 1 & 0 & 0 & 0 \ 0 & 0 & 1 & 1 & 0 \ 0 & 0 & 0 & 2 & 0 \ 0 & 0 & 0 & 0 & 2end{pmatrix}.$$It is diagonalizable: for instance, the vectors $(1,0,0,0,0)$, $(0,1,0,0,0)$, and $(0,0,1,0,0)$ are eigenvalues with eigenvalue $1$ and $(0,0,1,1,0)$ and $(0,0,0,0,1)$ are eigenvectors with eigenvalue $2$.
answered Apr 18 at 23:34
José Carlos Santos
148k22117218
Ah I made mistake. I forgot c = 0.
– Goldname
Apr 19 at 1:11
add a comment |
Why do you think that $b$ must be $0$? That's not true. Take, for instance this case:$$A=begin{pmatrix}1 & 0 & 0 & 0 & 0 \ 0 & 1 & 0 & 0 & 0 \ 0 & 0 & 1 & 1 & 0 \ 0 & 0 & 0 & 2 & 0 \ 0 & 0 & 0 & 0 & 2end{pmatrix}.$$It is diagonalizable: for instance, the vectors $(1,0,0,0,0)$, $(0,1,0,0,0)$, and $(0,0,1,0,0)$ are eigenvalues with eigenvalue $1$ and $(0,0,1,1,0)$ and $(0,0,0,0,1)$ are eigenvectors with eigenvalue $2$.
answered Apr 18 at 23:34
José Carlos Santos
148k22117218
Ah I made mistake. I forgot c = 0.
– Goldname
Apr 19 at 1:11
add a comment |
Why do you think that $b$ must be $0$? That's not true. Take, for instance this case:$$A=begin{pmatrix}1 & 0 & 0 & 0 & 0 \ 0 & 1 & 0 & 0 & 0 \ 0 & 0 & 1 & 1 & 0 \ 0 & 0 & 0 & 2 & 0 \ 0 & 0 & 0 & 0 & 2end{pmatrix}.$$It is diagonalizable: for instance, the vectors $(1,0,0,0,0)$, $(0,1,0,0,0)$, and $(0,0,1,0,0)$ are eigenvalues with eigenvalue $1$ and $(0,0,1,1,0)$ and $(0,0,0,0,1)$ are eigenvectors with eigenvalue $2$.
answered Apr 18 at 23:34
José Carlos Santos
148k22117218
Why do you think that $b$ must be $0$? That's not true. Take, for instance this case:$$A=begin{pmatrix}1 & 0 & 0 & 0 & 0 \ 0 & 1 & 0 & 0 & 0 \ 0 & 0 & 1 & 1 & 0 \ 0 & 0 & 0 & 2 & 0 \ 0 & 0 & 0 & 0 & 2end{pmatrix}.$$It is diagonalizable: for instance, the vectors $(1,0,0,0,0)$, $(0,1,0,0,0)$, and $(0,0,1,0,0)$ are eigenvalues with eigenvalue $1$ and $(0,0,1,1,0)$ and $(0,0,0,0,1)$ are eigenvectors with eigenvalue $2$.
answered Apr 18 at 23:34
José Carlos Santos
148k22117218
answered Apr 18 at 23:34
José Carlos Santos
148k22117218
answered Apr 18 at 23:34
José Carlos Santos
148k22117218
answered Apr 18 at 23:34
José Carlos Santos
148k22117218
148k22117218
Ah I made mistake. I forgot c = 0.
– Goldname
Apr 19 at 1:11
add a comment |
Ah I made mistake. I forgot c = 0.
– Goldname
Apr 19 at 1:11
Ah I made mistake. I forgot c = 0.
– Goldname
Apr 19 at 1:11
Ah I made mistake. I forgot c = 0.
– Goldname
Apr 19 at 1:11
add a comment |
With the given constraints on $a$ and $c$, $det A=b^2$, so its kernel is always trivial for $bne0$. If instead by the first “this matrix” you mean $$A-bI = begin{bmatrix}1&0&0&0&0\0&1&0&0&0\0&0&1&a&0\0&0&0&0&0\0&0&0&0&0end{bmatrix},$$ its null space is quite clearly two-dimensional regardless of the value of $a$.
answered Apr 18 at 23:50
amd
29k21050
add a comment |
With the given constraints on $a$ and $c$, $det A=b^2$, so its kernel is always trivial for $bne0$. If instead by the first “this matrix” you mean $$A-bI = begin{bmatrix}1&0&0&0&0\0&1&0&0&0\0&0&1&a&0\0&0&0&0&0\0&0&0&0&0end{bmatrix},$$ its null space is quite clearly two-dimensional regardless of the value of $a$.
answered Apr 18 at 23:50
amd
29k21050
add a comment |
With the given constraints on $a$ and $c$, $det A=b^2$, so its kernel is always trivial for $bne0$. If instead by the first “this matrix” you mean $$A-bI = begin{bmatrix}1&0&0&0&0\0&1&0&0&0\0&0&1&a&0\0&0&0&0&0\0&0&0&0&0end{bmatrix},$$ its null space is quite clearly two-dimensional regardless of the value of $a$.
answered Apr 18 at 23:50
amd
29k21050
With the given constraints on $a$ and $c$, $det A=b^2$, so its kernel is always trivial for $bne0$. If instead by the first “this matrix” you mean $$A-bI = begin{bmatrix}1&0&0&0&0\0&1&0&0&0\0&0&1&a&0\0&0&0&0&0\0&0&0&0&0end{bmatrix},$$ its null space is quite clearly two-dimensional regardless of the value of $a$.
answered Apr 18 at 23:50
amd
29k21050
answered Apr 18 at 23:50
amd
29k21050
answered Apr 18 at 23:50
amd
29k21050
answered Apr 18 at 23:50
amd
29k21050
29k21050
add a comment |
add a comment |
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