Mixing
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Proving $dim(E_0) geq n - k$
$begingroup$
I found a question from an old exam which I am not really able to wrap my head around. The questions states:
Given $k<n$ and $v_1, v_2, ..., v_k in mathbb{R} ^n$ non-zero vectors, orthogonal to the standard dot product on $mathbb{R}^n$, where we consider the vectors of $mathbb{R}^n$ as column vectors. Given that $lambda_1, lambda_2, ..., lambda_k in mathbb{R}$ and
$$A = lambda_1 v_1 v_1^T + lambda_2 v_2 v_2^T + ... + lambda_k v_k v_k^T in mathbb{R}^{n times n}$$
Part A asks to prove that $v_1, v_2, ..., v_k $ are eigenvectors of $A$. I managed to prove this by showing that $$Av_i = left(sum_{j=1}^klambda_jv_jv_j^Tright)v_i=sum_{j=1}^klambda_jv_j(v_j^Tv_i)=lambda_iv_i|| v_i||^2=left(lambda_i|| v_i||^2right)v_i$$ $forall i = 1, 2, ..., k$
Part B asks to prove that $dim(E_0) geq n - k$, where $E_0$ the eigenspace is for the eigenvalue $0$. I don't really have a direct idea of how to get started with this part.
Part C asks to prove that $A$ is diagonalisable and give an orthogonal basis of $A$. Again I am not really sure how to get started with this.
Any help is appreciated.
linear-algebra matrices eigenvalues-eigenvectors orthogonality diagonalization
asked Jan 22 at 16:41
Elliot SElliot S
657
$endgroup$
add a comment |
$begingroup$
I found a question from an old exam which I am not really able to wrap my head around. The questions states:
Given $k<n$ and $v_1, v_2, ..., v_k in mathbb{R} ^n$ non-zero vectors, orthogonal to the standard dot product on $mathbb{R}^n$, where we consider the vectors of $mathbb{R}^n$ as column vectors. Given that $lambda_1, lambda_2, ..., lambda_k in mathbb{R}$ and
$$A = lambda_1 v_1 v_1^T + lambda_2 v_2 v_2^T + ... + lambda_k v_k v_k^T in mathbb{R}^{n times n}$$
Part A asks to prove that $v_1, v_2, ..., v_k $ are eigenvectors of $A$. I managed to prove this by showing that $$Av_i = left(sum_{j=1}^klambda_jv_jv_j^Tright)v_i=sum_{j=1}^klambda_jv_j(v_j^Tv_i)=lambda_iv_i|| v_i||^2=left(lambda_i|| v_i||^2right)v_i$$ $forall i = 1, 2, ..., k$
Part B asks to prove that $dim(E_0) geq n - k$, where $E_0$ the eigenspace is for the eigenvalue $0$. I don't really have a direct idea of how to get started with this part.
Part C asks to prove that $A$ is diagonalisable and give an orthogonal basis of $A$. Again I am not really sure how to get started with this.
Any help is appreciated.
linear-algebra matrices eigenvalues-eigenvectors orthogonality diagonalization
asked Jan 22 at 16:41
Elliot SElliot S
657
$endgroup$
$begingroup$
I think $Bbb R^{2times2}$ should be $Bbb R^{ntimes n}$, which presumably denotes the space of $ntimes n$ matrices.
$endgroup$
– Marc van Leeuwen
Jan 22 at 17:12
add a comment |
$begingroup$
I found a question from an old exam which I am not really able to wrap my head around. The questions states:
Given $k<n$ and $v_1, v_2, ..., v_k in mathbb{R} ^n$ non-zero vectors, orthogonal to the standard dot product on $mathbb{R}^n$, where we consider the vectors of $mathbb{R}^n$ as column vectors. Given that $lambda_1, lambda_2, ..., lambda_k in mathbb{R}$ and
$$A = lambda_1 v_1 v_1^T + lambda_2 v_2 v_2^T + ... + lambda_k v_k v_k^T in mathbb{R}^{n times n}$$
Part A asks to prove that $v_1, v_2, ..., v_k $ are eigenvectors of $A$. I managed to prove this by showing that $$Av_i = left(sum_{j=1}^klambda_jv_jv_j^Tright)v_i=sum_{j=1}^klambda_jv_j(v_j^Tv_i)=lambda_iv_i|| v_i||^2=left(lambda_i|| v_i||^2right)v_i$$ $forall i = 1, 2, ..., k$
Part B asks to prove that $dim(E_0) geq n - k$, where $E_0$ the eigenspace is for the eigenvalue $0$. I don't really have a direct idea of how to get started with this part.
Part C asks to prove that $A$ is diagonalisable and give an orthogonal basis of $A$. Again I am not really sure how to get started with this.
Any help is appreciated.
linear-algebra matrices eigenvalues-eigenvectors orthogonality diagonalization
asked Jan 22 at 16:41
Elliot SElliot S
657
$endgroup$
I found a question from an old exam which I am not really able to wrap my head around. The questions states:
Given $k<n$ and $v_1, v_2, ..., v_k in mathbb{R} ^n$ non-zero vectors, orthogonal to the standard dot product on $mathbb{R}^n$, where we consider the vectors of $mathbb{R}^n$ as column vectors. Given that $lambda_1, lambda_2, ..., lambda_k in mathbb{R}$ and
$$A = lambda_1 v_1 v_1^T + lambda_2 v_2 v_2^T + ... + lambda_k v_k v_k^T in mathbb{R}^{n times n}$$
Part A asks to prove that $v_1, v_2, ..., v_k $ are eigenvectors of $A$. I managed to prove this by showing that $$Av_i = left(sum_{j=1}^klambda_jv_jv_j^Tright)v_i=sum_{j=1}^klambda_jv_j(v_j^Tv_i)=lambda_iv_i|| v_i||^2=left(lambda_i|| v_i||^2right)v_i$$ $forall i = 1, 2, ..., k$
Part B asks to prove that $dim(E_0) geq n - k$, where $E_0$ the eigenspace is for the eigenvalue $0$. I don't really have a direct idea of how to get started with this part.
Part C asks to prove that $A$ is diagonalisable and give an orthogonal basis of $A$. Again I am not really sure how to get started with this.
Any help is appreciated.
linear-algebra matrices eigenvalues-eigenvectors orthogonality diagonalization
linear-algebra matrices eigenvalues-eigenvectors orthogonality diagonalization
asked Jan 22 at 16:41
Elliot SElliot S
657
asked Jan 22 at 16:41
Elliot SElliot S
657
edited Jan 23 at 8:49
Elliot S
asked Jan 22 at 16:41
Elliot SElliot S
657
asked Jan 22 at 16:41
Elliot SElliot S
657
asked Jan 22 at 16:41
Elliot SElliot S
657
657
$begingroup$
I think $Bbb R^{2times2}$ should be $Bbb R^{ntimes n}$, which presumably denotes the space of $ntimes n$ matrices.
$endgroup$
– Marc van Leeuwen
Jan 22 at 17:12
add a comment |
$begingroup$
I think $Bbb R^{2times2}$ should be $Bbb R^{ntimes n}$, which presumably denotes the space of $ntimes n$ matrices.
$endgroup$
– Marc van Leeuwen
Jan 22 at 17:12
$begingroup$
I think $Bbb R^{2times2}$ should be $Bbb R^{ntimes n}$, which presumably denotes the space of $ntimes n$ matrices.
$endgroup$
– Marc van Leeuwen
Jan 22 at 17:12
$begingroup$
I think $Bbb R^{2times2}$ should be $Bbb R^{ntimes n}$, which presumably denotes the space of $ntimes n$ matrices.
$endgroup$
– Marc van Leeuwen
Jan 22 at 17:12
add a comment |
2 Answers
2
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For Part B, you can find an orthogonal system $$v_{k+1},v_{k+2},ldots, v_n ,$$ which is orthogonal to ${v_1,v_2,ldots, v_k}$ (e.g. by Gram-Schmidt). We can see $Av_{i}=0$ for $k<ile n$. From this, deduce that $dim E_0ge n-k$.
Combining the Part A and B, we can see immediately there exists an orthogonal basis consisting of eigenvectors of $A$.
Note: Spectral decomposition theorem says every symmetric real matrix is of the same form as that of $A$.
answered Jan 22 at 16:55
SongSong
16.9k21145
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add a comment |
$begingroup$
Part B: observe that "the eigenspace for the eigenvalue $0$" is just a fancy way of saying "null space". Both are the set of all vectors $v$ for which $Av=0$. I'm assuming you're familiar with computing null spaces?
Part C: a matrix is diagonalizable if the sum of dimensions of its eigenspaces equals the number of columns. For example, a $3times 3$ matrix will be diagonalizable if it has two eigenvalues, and the corresponding dimensions of the eigenspaces are $1$ and $2$. If we know that a matrix is diagonalizable, diagonalizing it amounts to finding all of the eigenvalue/eigenvector pairs. You can then use Gram-Schmidt to get an orthogonal basis.
answered Jan 22 at 16:49
pwerthpwerth
3,265417
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2 Answers
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active
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2 Answers
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active
oldest
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$begingroup$
For Part B, you can find an orthogonal system $$v_{k+1},v_{k+2},ldots, v_n ,$$ which is orthogonal to ${v_1,v_2,ldots, v_k}$ (e.g. by Gram-Schmidt). We can see $Av_{i}=0$ for $k<ile n$. From this, deduce that $dim E_0ge n-k$.
Combining the Part A and B, we can see immediately there exists an orthogonal basis consisting of eigenvectors of $A$.
Note: Spectral decomposition theorem says every symmetric real matrix is of the same form as that of $A$.
answered Jan 22 at 16:55
SongSong
16.9k21145
$endgroup$
add a comment |
$begingroup$
For Part B, you can find an orthogonal system $$v_{k+1},v_{k+2},ldots, v_n ,$$ which is orthogonal to ${v_1,v_2,ldots, v_k}$ (e.g. by Gram-Schmidt). We can see $Av_{i}=0$ for $k<ile n$. From this, deduce that $dim E_0ge n-k$.
Combining the Part A and B, we can see immediately there exists an orthogonal basis consisting of eigenvectors of $A$.
Note: Spectral decomposition theorem says every symmetric real matrix is of the same form as that of $A$.
answered Jan 22 at 16:55
SongSong
16.9k21145
$endgroup$
add a comment |
$begingroup$
For Part B, you can find an orthogonal system $$v_{k+1},v_{k+2},ldots, v_n ,$$ which is orthogonal to ${v_1,v_2,ldots, v_k}$ (e.g. by Gram-Schmidt). We can see $Av_{i}=0$ for $k<ile n$. From this, deduce that $dim E_0ge n-k$.
Combining the Part A and B, we can see immediately there exists an orthogonal basis consisting of eigenvectors of $A$.
Note: Spectral decomposition theorem says every symmetric real matrix is of the same form as that of $A$.
answered Jan 22 at 16:55
SongSong
16.9k21145
$endgroup$
For Part B, you can find an orthogonal system $$v_{k+1},v_{k+2},ldots, v_n ,$$ which is orthogonal to ${v_1,v_2,ldots, v_k}$ (e.g. by Gram-Schmidt). We can see $Av_{i}=0$ for $k<ile n$. From this, deduce that $dim E_0ge n-k$.
Combining the Part A and B, we can see immediately there exists an orthogonal basis consisting of eigenvectors of $A$.
Note: Spectral decomposition theorem says every symmetric real matrix is of the same form as that of $A$.
answered Jan 22 at 16:55
SongSong
16.9k21145
edited Jan 22 at 17:00
answered Jan 22 at 16:55
SongSong
16.9k21145
answered Jan 22 at 16:55
SongSong
16.9k21145
answered Jan 22 at 16:55
SongSong
16.9k21145
16.9k21145
add a comment |
add a comment |
$begingroup$
Part B: observe that "the eigenspace for the eigenvalue $0$" is just a fancy way of saying "null space". Both are the set of all vectors $v$ for which $Av=0$. I'm assuming you're familiar with computing null spaces?
Part C: a matrix is diagonalizable if the sum of dimensions of its eigenspaces equals the number of columns. For example, a $3times 3$ matrix will be diagonalizable if it has two eigenvalues, and the corresponding dimensions of the eigenspaces are $1$ and $2$. If we know that a matrix is diagonalizable, diagonalizing it amounts to finding all of the eigenvalue/eigenvector pairs. You can then use Gram-Schmidt to get an orthogonal basis.
answered Jan 22 at 16:49
pwerthpwerth
3,265417
$endgroup$
add a comment |
$begingroup$
Part B: observe that "the eigenspace for the eigenvalue $0$" is just a fancy way of saying "null space". Both are the set of all vectors $v$ for which $Av=0$. I'm assuming you're familiar with computing null spaces?
Part C: a matrix is diagonalizable if the sum of dimensions of its eigenspaces equals the number of columns. For example, a $3times 3$ matrix will be diagonalizable if it has two eigenvalues, and the corresponding dimensions of the eigenspaces are $1$ and $2$. If we know that a matrix is diagonalizable, diagonalizing it amounts to finding all of the eigenvalue/eigenvector pairs. You can then use Gram-Schmidt to get an orthogonal basis.
answered Jan 22 at 16:49
pwerthpwerth
3,265417
$endgroup$
add a comment |
$begingroup$
Part B: observe that "the eigenspace for the eigenvalue $0$" is just a fancy way of saying "null space". Both are the set of all vectors $v$ for which $Av=0$. I'm assuming you're familiar with computing null spaces?
Part C: a matrix is diagonalizable if the sum of dimensions of its eigenspaces equals the number of columns. For example, a $3times 3$ matrix will be diagonalizable if it has two eigenvalues, and the corresponding dimensions of the eigenspaces are $1$ and $2$. If we know that a matrix is diagonalizable, diagonalizing it amounts to finding all of the eigenvalue/eigenvector pairs. You can then use Gram-Schmidt to get an orthogonal basis.
answered Jan 22 at 16:49
pwerthpwerth
3,265417
$endgroup$
Part B: observe that "the eigenspace for the eigenvalue $0$" is just a fancy way of saying "null space". Both are the set of all vectors $v$ for which $Av=0$. I'm assuming you're familiar with computing null spaces?
Part C: a matrix is diagonalizable if the sum of dimensions of its eigenspaces equals the number of columns. For example, a $3times 3$ matrix will be diagonalizable if it has two eigenvalues, and the corresponding dimensions of the eigenspaces are $1$ and $2$. If we know that a matrix is diagonalizable, diagonalizing it amounts to finding all of the eigenvalue/eigenvector pairs. You can then use Gram-Schmidt to get an orthogonal basis.
answered Jan 22 at 16:49
pwerthpwerth
3,265417
answered Jan 22 at 16:49
pwerthpwerth
3,265417
answered Jan 22 at 16:49
pwerthpwerth
3,265417
answered Jan 22 at 16:49
pwerthpwerth
3,265417
3,265417
add a comment |
add a comment |
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$begingroup$
I think $Bbb R^{2times2}$ should be $Bbb R^{ntimes n}$, which presumably denotes the space of $ntimes n$ matrices.
$endgroup$
– Marc van Leeuwen
Jan 22 at 17:12