Mixing
Linear independent eigenvectors and eigenvalues
I have T as a linear transformation from V to V over the field F, V has dimension n. T has the maximum number n distinct eigenvalues, then show that there exists a basis of V consisting of eigenvectors.
I know that if I let $v_1,...,v_r$ be eigenvectors belonging to distinct eigenvalues, then those vectors are linearly independent. Can I make a basis from these linearly independent vector and prove that it spans V?
Also, what will the matrix of T be in respect to this basis?
Thank you for any input!
linear-algebra matrices transformation eigenvalues-eigenvectors
edited Nov 20 '13 at 4:44
Mhenni Benghorbal
43.1k63574
asked Nov 20 '13 at 3:10
Akaichan
1,49421838
add a comment |
I have T as a linear transformation from V to V over the field F, V has dimension n. T has the maximum number n distinct eigenvalues, then show that there exists a basis of V consisting of eigenvectors.
I know that if I let $v_1,...,v_r$ be eigenvectors belonging to distinct eigenvalues, then those vectors are linearly independent. Can I make a basis from these linearly independent vector and prove that it spans V?
Also, what will the matrix of T be in respect to this basis?
Thank you for any input!
linear-algebra matrices transformation eigenvalues-eigenvectors
edited Nov 20 '13 at 4:44
Mhenni Benghorbal
43.1k63574
asked Nov 20 '13 at 3:10
Akaichan
1,49421838
1
If there are $n$ linearly independent vectors in $n$-dimensional space, then they must form a basis. To see what $T$ looks like, consider what $T x_k$ looks like in the basis of eigenvectors.
– copper.hat
Nov 20 '13 at 3:19
@Akaichan Do you still need help with this or is Copper.Hat's comment enough.
– Git Gud
Nov 25 '13 at 21:41
add a comment |
I have T as a linear transformation from V to V over the field F, V has dimension n. T has the maximum number n distinct eigenvalues, then show that there exists a basis of V consisting of eigenvectors.
I know that if I let $v_1,...,v_r$ be eigenvectors belonging to distinct eigenvalues, then those vectors are linearly independent. Can I make a basis from these linearly independent vector and prove that it spans V?
Also, what will the matrix of T be in respect to this basis?
Thank you for any input!
linear-algebra matrices transformation eigenvalues-eigenvectors
edited Nov 20 '13 at 4:44
Mhenni Benghorbal
43.1k63574
asked Nov 20 '13 at 3:10
Akaichan
1,49421838
I have T as a linear transformation from V to V over the field F, V has dimension n. T has the maximum number n distinct eigenvalues, then show that there exists a basis of V consisting of eigenvectors.
I know that if I let $v_1,...,v_r$ be eigenvectors belonging to distinct eigenvalues, then those vectors are linearly independent. Can I make a basis from these linearly independent vector and prove that it spans V?
Also, what will the matrix of T be in respect to this basis?
Thank you for any input!
linear-algebra matrices transformation eigenvalues-eigenvectors
linear-algebra matrices transformation eigenvalues-eigenvectors
edited Nov 20 '13 at 4:44
Mhenni Benghorbal
43.1k63574
asked Nov 20 '13 at 3:10
Akaichan
1,49421838
edited Nov 20 '13 at 4:44
Mhenni Benghorbal
43.1k63574
asked Nov 20 '13 at 3:10
Akaichan
1,49421838
edited Nov 20 '13 at 4:44
Mhenni Benghorbal
43.1k63574
edited Nov 20 '13 at 4:44
Mhenni Benghorbal
43.1k63574
edited Nov 20 '13 at 4:44
Mhenni Benghorbal
43.1k63574
43.1k63574
asked Nov 20 '13 at 3:10
Akaichan
1,49421838
asked Nov 20 '13 at 3:10
Akaichan
1,49421838
asked Nov 20 '13 at 3:10
Akaichan
1,49421838
1,49421838
1
If there are $n$ linearly independent vectors in $n$-dimensional space, then they must form a basis. To see what $T$ looks like, consider what $T x_k$ looks like in the basis of eigenvectors.
– copper.hat
Nov 20 '13 at 3:19
@Akaichan Do you still need help with this or is Copper.Hat's comment enough.
– Git Gud
Nov 25 '13 at 21:41
add a comment |
1
If there are $n$ linearly independent vectors in $n$-dimensional space, then they must form a basis. To see what $T$ looks like, consider what $T x_k$ looks like in the basis of eigenvectors.
– copper.hat
Nov 20 '13 at 3:19
@Akaichan Do you still need help with this or is Copper.Hat's comment enough.
– Git Gud
Nov 25 '13 at 21:41
1
1
If there are $n$ linearly independent vectors in $n$-dimensional space, then they must form a basis. To see what $T$ looks like, consider what $T x_k$ looks like in the basis of eigenvectors.
– copper.hat
Nov 20 '13 at 3:19
If there are $n$ linearly independent vectors in $n$-dimensional space, then they must form a basis. To see what $T$ looks like, consider what $T x_k$ looks like in the basis of eigenvectors.
– copper.hat
Nov 20 '13 at 3:19
@Akaichan Do you still need help with this or is Copper.Hat's comment enough.
– Git Gud
Nov 25 '13 at 21:41
@Akaichan Do you still need help with this or is Copper.Hat's comment enough.
– Git Gud
Nov 25 '13 at 21:41
add a comment |
1 Answer
1
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By definition, a basis for $V$ is a linearly-independent set of vectors in $V$ that spans the space $V,$ and the dimension of a finite-dimensional vector space is the number of elements in a basis for $V,$ so we know that any basis for $V$ must contain exactly $n$ linearly-independent vectors. We also know (see Theorem 5 on page 45 of Hoffman and Kunze's Linear Algebra) that every linearly-independent subset of $V$ is part of a basis for $V.$ You already know that the $n$ eigenvectors are linearly independent, so it follows that they form a basis for $V.$
To find the matrix of $T$ with respect your ordered basis $mathscr B$ of eigenvectors, we use the fact that the ith column of that matrix is given by $[Tv_i]_{mathscr B}$ where $[ cdot ]_{mathscr B}$ denotes the coordinate matrix with respect to $mathscr B.$ We therefore compute
$$[Tv_i]_{mathscr B} = [lambda_i v_i]_{mathscr B} = begin{bmatrix} 0\ vdots\ 0\ lambda_i\ 0\ vdots\ 0end{bmatrix}$$
where $lambda_i$ is the eigenvalue associated with eigenvector $v_i.$ Thus, the matrix of $T$ is a diagonal matrix with the eigenvalues in the diagonal entries ordered by corresponding eigenvectors:
$$begin{bmatrix}
lambda_1 & 0 & cdots & 0\
0 & lambda_2 & cdots & 0\
vdots & vdots & ddots & vdots\
0 & 0 & cdots & lambda_nend{bmatrix}.$$
answered Nov 20 '18 at 17:34
Maurice P
1,3901732
add a comment |
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By definition, a basis for $V$ is a linearly-independent set of vectors in $V$ that spans the space $V,$ and the dimension of a finite-dimensional vector space is the number of elements in a basis for $V,$ so we know that any basis for $V$ must contain exactly $n$ linearly-independent vectors. We also know (see Theorem 5 on page 45 of Hoffman and Kunze's Linear Algebra) that every linearly-independent subset of $V$ is part of a basis for $V.$ You already know that the $n$ eigenvectors are linearly independent, so it follows that they form a basis for $V.$
To find the matrix of $T$ with respect your ordered basis $mathscr B$ of eigenvectors, we use the fact that the ith column of that matrix is given by $[Tv_i]_{mathscr B}$ where $[ cdot ]_{mathscr B}$ denotes the coordinate matrix with respect to $mathscr B.$ We therefore compute
$$[Tv_i]_{mathscr B} = [lambda_i v_i]_{mathscr B} = begin{bmatrix} 0\ vdots\ 0\ lambda_i\ 0\ vdots\ 0end{bmatrix}$$
where $lambda_i$ is the eigenvalue associated with eigenvector $v_i.$ Thus, the matrix of $T$ is a diagonal matrix with the eigenvalues in the diagonal entries ordered by corresponding eigenvectors:
$$begin{bmatrix}
lambda_1 & 0 & cdots & 0\
0 & lambda_2 & cdots & 0\
vdots & vdots & ddots & vdots\
0 & 0 & cdots & lambda_nend{bmatrix}.$$
answered Nov 20 '18 at 17:34
Maurice P
1,3901732
add a comment |
By definition, a basis for $V$ is a linearly-independent set of vectors in $V$ that spans the space $V,$ and the dimension of a finite-dimensional vector space is the number of elements in a basis for $V,$ so we know that any basis for $V$ must contain exactly $n$ linearly-independent vectors. We also know (see Theorem 5 on page 45 of Hoffman and Kunze's Linear Algebra) that every linearly-independent subset of $V$ is part of a basis for $V.$ You already know that the $n$ eigenvectors are linearly independent, so it follows that they form a basis for $V.$
To find the matrix of $T$ with respect your ordered basis $mathscr B$ of eigenvectors, we use the fact that the ith column of that matrix is given by $[Tv_i]_{mathscr B}$ where $[ cdot ]_{mathscr B}$ denotes the coordinate matrix with respect to $mathscr B.$ We therefore compute
$$[Tv_i]_{mathscr B} = [lambda_i v_i]_{mathscr B} = begin{bmatrix} 0\ vdots\ 0\ lambda_i\ 0\ vdots\ 0end{bmatrix}$$
where $lambda_i$ is the eigenvalue associated with eigenvector $v_i.$ Thus, the matrix of $T$ is a diagonal matrix with the eigenvalues in the diagonal entries ordered by corresponding eigenvectors:
$$begin{bmatrix}
lambda_1 & 0 & cdots & 0\
0 & lambda_2 & cdots & 0\
vdots & vdots & ddots & vdots\
0 & 0 & cdots & lambda_nend{bmatrix}.$$
answered Nov 20 '18 at 17:34
Maurice P
1,3901732
add a comment |
By definition, a basis for $V$ is a linearly-independent set of vectors in $V$ that spans the space $V,$ and the dimension of a finite-dimensional vector space is the number of elements in a basis for $V,$ so we know that any basis for $V$ must contain exactly $n$ linearly-independent vectors. We also know (see Theorem 5 on page 45 of Hoffman and Kunze's Linear Algebra) that every linearly-independent subset of $V$ is part of a basis for $V.$ You already know that the $n$ eigenvectors are linearly independent, so it follows that they form a basis for $V.$
To find the matrix of $T$ with respect your ordered basis $mathscr B$ of eigenvectors, we use the fact that the ith column of that matrix is given by $[Tv_i]_{mathscr B}$ where $[ cdot ]_{mathscr B}$ denotes the coordinate matrix with respect to $mathscr B.$ We therefore compute
$$[Tv_i]_{mathscr B} = [lambda_i v_i]_{mathscr B} = begin{bmatrix} 0\ vdots\ 0\ lambda_i\ 0\ vdots\ 0end{bmatrix}$$
where $lambda_i$ is the eigenvalue associated with eigenvector $v_i.$ Thus, the matrix of $T$ is a diagonal matrix with the eigenvalues in the diagonal entries ordered by corresponding eigenvectors:
$$begin{bmatrix}
lambda_1 & 0 & cdots & 0\
0 & lambda_2 & cdots & 0\
vdots & vdots & ddots & vdots\
0 & 0 & cdots & lambda_nend{bmatrix}.$$
answered Nov 20 '18 at 17:34
Maurice P
1,3901732
By definition, a basis for $V$ is a linearly-independent set of vectors in $V$ that spans the space $V,$ and the dimension of a finite-dimensional vector space is the number of elements in a basis for $V,$ so we know that any basis for $V$ must contain exactly $n$ linearly-independent vectors. We also know (see Theorem 5 on page 45 of Hoffman and Kunze's Linear Algebra) that every linearly-independent subset of $V$ is part of a basis for $V.$ You already know that the $n$ eigenvectors are linearly independent, so it follows that they form a basis for $V.$
To find the matrix of $T$ with respect your ordered basis $mathscr B$ of eigenvectors, we use the fact that the ith column of that matrix is given by $[Tv_i]_{mathscr B}$ where $[ cdot ]_{mathscr B}$ denotes the coordinate matrix with respect to $mathscr B.$ We therefore compute
$$[Tv_i]_{mathscr B} = [lambda_i v_i]_{mathscr B} = begin{bmatrix} 0\ vdots\ 0\ lambda_i\ 0\ vdots\ 0end{bmatrix}$$
where $lambda_i$ is the eigenvalue associated with eigenvector $v_i.$ Thus, the matrix of $T$ is a diagonal matrix with the eigenvalues in the diagonal entries ordered by corresponding eigenvectors:
$$begin{bmatrix}
lambda_1 & 0 & cdots & 0\
0 & lambda_2 & cdots & 0\
vdots & vdots & ddots & vdots\
0 & 0 & cdots & lambda_nend{bmatrix}.$$
answered Nov 20 '18 at 17:34
Maurice P
1,3901732
edited Nov 21 '18 at 17:19
answered Nov 20 '18 at 17:34
Maurice P
1,3901732
answered Nov 20 '18 at 17:34
Maurice P
1,3901732
answered Nov 20 '18 at 17:34
Maurice P
1,3901732
1,3901732
add a comment |
add a comment |
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If there are $n$ linearly independent vectors in $n$-dimensional space, then they must form a basis. To see what $T$ looks like, consider what $T x_k$ looks like in the basis of eigenvectors.
– copper.hat
Nov 20 '13 at 3:19
@Akaichan Do you still need help with this or is Copper.Hat's comment enough.
– Git Gud
Nov 25 '13 at 21:41