Finding Jordan basis of $5 times 5$ nilpotent matrix
$begingroup$
I have $5 times 5$ real matrix, which is nilpotent:
$$
A = begin{bmatrix}
-2 & 2 & 1 & 3 & -1 \
3 & -8 & -2 & -9 & 3 \
-2 &-8&0 & -6 & 2 \
-4 & 8 & 2 & 9 & -3 \
-4 & -4& 0 &-3 & 1
end{bmatrix}.
$$
I have successfully found the Jordan normal form, it is:
$$
J = begin{bmatrix}
0 & 0 & 0 & 0 & 0 \
1 & 0 & 0 & 0 & 0 \
0 & 0 & 0 & 0 & 0 \
0 & 0 & 1 & 0 & 0 \
0 & 0 & 0 & 1 & 0
end{bmatrix}.
$$
Now I am trying to find a Jordan basis.
First of all, I square $A$, obtaining
$$
A^2 = begin{bmatrix}
0 & 0 & 0 & 0 & 0 \
-2 & 2 & 1 & 3 & -1 \
-4 & 4 & 2 & 6 & -2 \
4 & -4 & -2 & -6 & 2 \
4 & -4 & -2 &-6 & 2
end{bmatrix}
$$
The cube equals zero: $A^3 = 0$. Now we see one column of $A^2$ basis. It is a third column : $$ vec{e}_1 = vec{z^{(2)}_1} = (0, 1, 2, -2, -2)^T $$ (by $vec{z^{(j)}_i}$ I am noting vectors, that I am looking for, where $i$ - number of vector in $R^j$, $R^j$ is a space induced by columns of matrix $(A - lambda E) ^ j$).
Next I need to find basis vectors of columns from $A$ and also to find $vec{z^{(1)}_1}$. Then, because of $A$ has dimension 3, we need to find $vec{z^{(1)}_2}$ also to complete the basis for $R^1$.
Already at this step I don't know, how to find vectors $vec{z^{(j)}_i}$
Please, help me. Also I would be grateful, if you could tell me all of the way to finding Jordan basis (untill the end, i.e. fifth vector).
linear-algebra matrices jordan-normal-form
edited Dec 8 '18 at 8:40
Eric Wofsey
182k13209337
asked Dec 7 '18 at 18:56
JuicerJuicer
11
$endgroup$
add a comment |
$begingroup$
I have $5 times 5$ real matrix, which is nilpotent:
$$
A = begin{bmatrix}
-2 & 2 & 1 & 3 & -1 \
3 & -8 & -2 & -9 & 3 \
-2 &-8&0 & -6 & 2 \
-4 & 8 & 2 & 9 & -3 \
-4 & -4& 0 &-3 & 1
end{bmatrix}.
$$
I have successfully found the Jordan normal form, it is:
$$
J = begin{bmatrix}
0 & 0 & 0 & 0 & 0 \
1 & 0 & 0 & 0 & 0 \
0 & 0 & 0 & 0 & 0 \
0 & 0 & 1 & 0 & 0 \
0 & 0 & 0 & 1 & 0
end{bmatrix}.
$$
Now I am trying to find a Jordan basis.
First of all, I square $A$, obtaining
$$
A^2 = begin{bmatrix}
0 & 0 & 0 & 0 & 0 \
-2 & 2 & 1 & 3 & -1 \
-4 & 4 & 2 & 6 & -2 \
4 & -4 & -2 & -6 & 2 \
4 & -4 & -2 &-6 & 2
end{bmatrix}
$$
The cube equals zero: $A^3 = 0$. Now we see one column of $A^2$ basis. It is a third column : $$ vec{e}_1 = vec{z^{(2)}_1} = (0, 1, 2, -2, -2)^T $$ (by $vec{z^{(j)}_i}$ I am noting vectors, that I am looking for, where $i$ - number of vector in $R^j$, $R^j$ is a space induced by columns of matrix $(A - lambda E) ^ j$).
Next I need to find basis vectors of columns from $A$ and also to find $vec{z^{(1)}_1}$. Then, because of $A$ has dimension 3, we need to find $vec{z^{(1)}_2}$ also to complete the basis for $R^1$.
Already at this step I don't know, how to find vectors $vec{z^{(j)}_i}$
Please, help me. Also I would be grateful, if you could tell me all of the way to finding Jordan basis (untill the end, i.e. fifth vector).
linear-algebra matrices jordan-normal-form
edited Dec 8 '18 at 8:40
Eric Wofsey
182k13209337
asked Dec 7 '18 at 18:56
JuicerJuicer
11
$endgroup$
$begingroup$
Title:"nilpotent matrix" (not nillpotent).
$endgroup$
– Dietrich Burde
Dec 7 '18 at 20:00
add a comment |
$begingroup$
I have $5 times 5$ real matrix, which is nilpotent:
$$
A = begin{bmatrix}
-2 & 2 & 1 & 3 & -1 \
3 & -8 & -2 & -9 & 3 \
-2 &-8&0 & -6 & 2 \
-4 & 8 & 2 & 9 & -3 \
-4 & -4& 0 &-3 & 1
end{bmatrix}.
$$
I have successfully found the Jordan normal form, it is:
$$
J = begin{bmatrix}
0 & 0 & 0 & 0 & 0 \
1 & 0 & 0 & 0 & 0 \
0 & 0 & 0 & 0 & 0 \
0 & 0 & 1 & 0 & 0 \
0 & 0 & 0 & 1 & 0
end{bmatrix}.
$$
Now I am trying to find a Jordan basis.
First of all, I square $A$, obtaining
$$
A^2 = begin{bmatrix}
0 & 0 & 0 & 0 & 0 \
-2 & 2 & 1 & 3 & -1 \
-4 & 4 & 2 & 6 & -2 \
4 & -4 & -2 & -6 & 2 \
4 & -4 & -2 &-6 & 2
end{bmatrix}
$$
The cube equals zero: $A^3 = 0$. Now we see one column of $A^2$ basis. It is a third column : $$ vec{e}_1 = vec{z^{(2)}_1} = (0, 1, 2, -2, -2)^T $$ (by $vec{z^{(j)}_i}$ I am noting vectors, that I am looking for, where $i$ - number of vector in $R^j$, $R^j$ is a space induced by columns of matrix $(A - lambda E) ^ j$).
Next I need to find basis vectors of columns from $A$ and also to find $vec{z^{(1)}_1}$. Then, because of $A$ has dimension 3, we need to find $vec{z^{(1)}_2}$ also to complete the basis for $R^1$.
Already at this step I don't know, how to find vectors $vec{z^{(j)}_i}$
Please, help me. Also I would be grateful, if you could tell me all of the way to finding Jordan basis (untill the end, i.e. fifth vector).
linear-algebra matrices jordan-normal-form
edited Dec 8 '18 at 8:40
Eric Wofsey
182k13209337
asked Dec 7 '18 at 18:56
JuicerJuicer
11
$endgroup$
I have $5 times 5$ real matrix, which is nilpotent:
$$
A = begin{bmatrix}
-2 & 2 & 1 & 3 & -1 \
3 & -8 & -2 & -9 & 3 \
-2 &-8&0 & -6 & 2 \
-4 & 8 & 2 & 9 & -3 \
-4 & -4& 0 &-3 & 1
end{bmatrix}.
$$
I have successfully found the Jordan normal form, it is:
$$
J = begin{bmatrix}
0 & 0 & 0 & 0 & 0 \
1 & 0 & 0 & 0 & 0 \
0 & 0 & 0 & 0 & 0 \
0 & 0 & 1 & 0 & 0 \
0 & 0 & 0 & 1 & 0
end{bmatrix}.
$$
Now I am trying to find a Jordan basis.
First of all, I square $A$, obtaining
$$
A^2 = begin{bmatrix}
0 & 0 & 0 & 0 & 0 \
-2 & 2 & 1 & 3 & -1 \
-4 & 4 & 2 & 6 & -2 \
4 & -4 & -2 & -6 & 2 \
4 & -4 & -2 &-6 & 2
end{bmatrix}
$$
The cube equals zero: $A^3 = 0$. Now we see one column of $A^2$ basis. It is a third column : $$ vec{e}_1 = vec{z^{(2)}_1} = (0, 1, 2, -2, -2)^T $$ (by $vec{z^{(j)}_i}$ I am noting vectors, that I am looking for, where $i$ - number of vector in $R^j$, $R^j$ is a space induced by columns of matrix $(A - lambda E) ^ j$).
Next I need to find basis vectors of columns from $A$ and also to find $vec{z^{(1)}_1}$. Then, because of $A$ has dimension 3, we need to find $vec{z^{(1)}_2}$ also to complete the basis for $R^1$.
Already at this step I don't know, how to find vectors $vec{z^{(j)}_i}$
Please, help me. Also I would be grateful, if you could tell me all of the way to finding Jordan basis (untill the end, i.e. fifth vector).
linear-algebra matrices jordan-normal-form
linear-algebra matrices jordan-normal-form
edited Dec 8 '18 at 8:40
Eric Wofsey
182k13209337
asked Dec 7 '18 at 18:56
JuicerJuicer
11
edited Dec 8 '18 at 8:40
Eric Wofsey
182k13209337
asked Dec 7 '18 at 18:56
JuicerJuicer
11
edited Dec 8 '18 at 8:40
Eric Wofsey
182k13209337
edited Dec 8 '18 at 8:40
Eric Wofsey
182k13209337
edited Dec 8 '18 at 8:40
Eric Wofsey
182k13209337
182k13209337
asked Dec 7 '18 at 18:56
JuicerJuicer
11
asked Dec 7 '18 at 18:56
JuicerJuicer
11
asked Dec 7 '18 at 18:56
JuicerJuicer
11
11
$begingroup$
Title:"nilpotent matrix" (not nillpotent).
$endgroup$
– Dietrich Burde
Dec 7 '18 at 20:00
add a comment |
$begingroup$
Title:"nilpotent matrix" (not nillpotent).
$endgroup$
– Dietrich Burde
Dec 7 '18 at 20:00
$begingroup$
Title:"nilpotent matrix" (not nillpotent).
$endgroup$
– Dietrich Burde
Dec 7 '18 at 20:00
$begingroup$
Title:"nilpotent matrix" (not nillpotent).
$endgroup$
– Dietrich Burde
Dec 7 '18 at 20:00
add a comment |
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$begingroup$
Title:"nilpotent matrix" (not nillpotent).
$endgroup$
– Dietrich Burde
Dec 7 '18 at 20:00