Determine dimension of set of matrices given that each matrix has same eigenvalue 0 and eigenvector
$begingroup$
Let V be the set of all $A_{3×3}$ matrices, where all of them have eigenvalue 0 and corresponding eigenvector $y=(1,2,3)^T$.
Assuming we have showed that V is a vector space, determine the dimension of V.
From the definition we know that:
$x$ and $lambda$ are eigenvector and eigenvalue iff $(A-lambda I)x=0$ .
So for each matrix $A_i$ in our set$$
(A_i-0)
begin{pmatrix}
1 \
2 \
3 \
end{pmatrix}
=0
$$
$$
A_i
begin{pmatrix}
1 \
2 \
3 \
end{pmatrix}
=0
$$
$$
begin{pmatrix}
1 \
2 \
3 \
end{pmatrix}
in N(A_i)
$$
From this we find that $Nullity(A_i)geq1 Rightarrow $
$Rank(A_i) = n-Nullity(A_i)=3-1=2 Rightarrow $
$Rank(A_i)=Dim(C(A_i)) leq 2$
However, I am not sure how can I proceed from here?
1) How can determine the dimension of the matrix $A_i$ for sure?
2) How is that connected to the dimension of entire vector space formed by such matrices?
linear-algebra vector-spaces eigenvalues-eigenvectors
$endgroup$
add a comment |
$begingroup$
Let V be the set of all $A_{3×3}$ matrices, where all of them have eigenvalue 0 and corresponding eigenvector $y=(1,2,3)^T$.
Assuming we have showed that V is a vector space, determine the dimension of V.
From the definition we know that:
$x$ and $lambda$ are eigenvector and eigenvalue iff $(A-lambda I)x=0$ .
So for each matrix $A_i$ in our set$$
(A_i-0)
begin{pmatrix}
1 \
2 \
3 \
end{pmatrix}
=0
$$
$$
A_i
begin{pmatrix}
1 \
2 \
3 \
end{pmatrix}
=0
$$
$$
begin{pmatrix}
1 \
2 \
3 \
end{pmatrix}
in N(A_i)
$$
From this we find that $Nullity(A_i)geq1 Rightarrow $
$Rank(A_i) = n-Nullity(A_i)=3-1=2 Rightarrow $
$Rank(A_i)=Dim(C(A_i)) leq 2$
However, I am not sure how can I proceed from here?
1) How can determine the dimension of the matrix $A_i$ for sure?
2) How is that connected to the dimension of entire vector space formed by such matrices?
linear-algebra vector-spaces eigenvalues-eigenvectors
$endgroup$
add a comment |
$begingroup$
Let V be the set of all $A_{3×3}$ matrices, where all of them have eigenvalue 0 and corresponding eigenvector $y=(1,2,3)^T$.
Assuming we have showed that V is a vector space, determine the dimension of V.
From the definition we know that:
$x$ and $lambda$ are eigenvector and eigenvalue iff $(A-lambda I)x=0$ .
So for each matrix $A_i$ in our set$$
(A_i-0)
begin{pmatrix}
1 \
2 \
3 \
end{pmatrix}
=0
$$
$$
A_i
begin{pmatrix}
1 \
2 \
3 \
end{pmatrix}
=0
$$
$$
begin{pmatrix}
1 \
2 \
3 \
end{pmatrix}
in N(A_i)
$$
From this we find that $Nullity(A_i)geq1 Rightarrow $
$Rank(A_i) = n-Nullity(A_i)=3-1=2 Rightarrow $
$Rank(A_i)=Dim(C(A_i)) leq 2$
However, I am not sure how can I proceed from here?
1) How can determine the dimension of the matrix $A_i$ for sure?
2) How is that connected to the dimension of entire vector space formed by such matrices?
linear-algebra vector-spaces eigenvalues-eigenvectors
$endgroup$
Let V be the set of all $A_{3×3}$ matrices, where all of them have eigenvalue 0 and corresponding eigenvector $y=(1,2,3)^T$.
Assuming we have showed that V is a vector space, determine the dimension of V.
From the definition we know that:
$x$ and $lambda$ are eigenvector and eigenvalue iff $(A-lambda I)x=0$ .
So for each matrix $A_i$ in our set$$
(A_i-0)
begin{pmatrix}
1 \
2 \
3 \
end{pmatrix}
=0
$$
$$
A_i
begin{pmatrix}
1 \
2 \
3 \
end{pmatrix}
=0
$$
$$
begin{pmatrix}
1 \
2 \
3 \
end{pmatrix}
in N(A_i)
$$
From this we find that $Nullity(A_i)geq1 Rightarrow $
$Rank(A_i) = n-Nullity(A_i)=3-1=2 Rightarrow $
$Rank(A_i)=Dim(C(A_i)) leq 2$
However, I am not sure how can I proceed from here?
1) How can determine the dimension of the matrix $A_i$ for sure?
2) How is that connected to the dimension of entire vector space formed by such matrices?
linear-algebra vector-spaces eigenvalues-eigenvectors
linear-algebra vector-spaces eigenvalues-eigenvectors
asked Jan 15 at 19:38
Jill WhiteJill White
133
133
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