Determine the $|A|_2$ vector-induced norm
$begingroup$
This seems to be a very easy exercise, but as I never had previous contact with vector induced norm of a matrix, I am getting stuck...
Given the matrix:
$$A=
begin{bmatrix}
2 & 0\
0 & 1\
end{bmatrix}
$$
I am supposed to compute its vector-induced $|A|_2$ norm, defined as:
$$|A|_2=sup_{vneq 0}frac{|Av|_2}{|v|_2}$$
My attempt so far was:
Define a vector $v=(v_1,v_2)$ so that I end up with:
$$|A|_2=sup_{vneq 0}frac{|Av|_2}{|v|_2}=sup_{vneq 0}sqrt{frac{4v_{1}^{2}+v_{2}^{2}}{v_{1}^{2}+v_{2}^{2}}}$$
It seems reasonable to choose a vector $v$ such that $|v|_2=1$, then this would result in a problem where I want to find:
$$|A|_2=sup_{|v|_2=1}sqrt{4v_{1}^{2}+v_{2}^{2}}$$
Which kind of seems like an optimization problem where the constraint is:
$$sqrt{v_{1}^{2}+v_{2}^{2}}=1$$
And this is where I get stuck!
Is this a correct way to solve this? If yes, could you help me finish this line of thought? Would Lagrange Multiplier do the job?
I really would like to solve it "the hard way" before using other properties/definitions such as:
$$|A|_2=sigma_{max}(A)$$
Thanks in advance!
linear-algebra optimization vectors norm
$endgroup$
add a comment |
$begingroup$
This seems to be a very easy exercise, but as I never had previous contact with vector induced norm of a matrix, I am getting stuck...
Given the matrix:
$$A=
begin{bmatrix}
2 & 0\
0 & 1\
end{bmatrix}
$$
I am supposed to compute its vector-induced $|A|_2$ norm, defined as:
$$|A|_2=sup_{vneq 0}frac{|Av|_2}{|v|_2}$$
My attempt so far was:
Define a vector $v=(v_1,v_2)$ so that I end up with:
$$|A|_2=sup_{vneq 0}frac{|Av|_2}{|v|_2}=sup_{vneq 0}sqrt{frac{4v_{1}^{2}+v_{2}^{2}}{v_{1}^{2}+v_{2}^{2}}}$$
It seems reasonable to choose a vector $v$ such that $|v|_2=1$, then this would result in a problem where I want to find:
$$|A|_2=sup_{|v|_2=1}sqrt{4v_{1}^{2}+v_{2}^{2}}$$
Which kind of seems like an optimization problem where the constraint is:
$$sqrt{v_{1}^{2}+v_{2}^{2}}=1$$
And this is where I get stuck!
Is this a correct way to solve this? If yes, could you help me finish this line of thought? Would Lagrange Multiplier do the job?
I really would like to solve it "the hard way" before using other properties/definitions such as:
$$|A|_2=sigma_{max}(A)$$
Thanks in advance!
linear-algebra optimization vectors norm
$endgroup$
add a comment |
$begingroup$
This seems to be a very easy exercise, but as I never had previous contact with vector induced norm of a matrix, I am getting stuck...
Given the matrix:
$$A=
begin{bmatrix}
2 & 0\
0 & 1\
end{bmatrix}
$$
I am supposed to compute its vector-induced $|A|_2$ norm, defined as:
$$|A|_2=sup_{vneq 0}frac{|Av|_2}{|v|_2}$$
My attempt so far was:
Define a vector $v=(v_1,v_2)$ so that I end up with:
$$|A|_2=sup_{vneq 0}frac{|Av|_2}{|v|_2}=sup_{vneq 0}sqrt{frac{4v_{1}^{2}+v_{2}^{2}}{v_{1}^{2}+v_{2}^{2}}}$$
It seems reasonable to choose a vector $v$ such that $|v|_2=1$, then this would result in a problem where I want to find:
$$|A|_2=sup_{|v|_2=1}sqrt{4v_{1}^{2}+v_{2}^{2}}$$
Which kind of seems like an optimization problem where the constraint is:
$$sqrt{v_{1}^{2}+v_{2}^{2}}=1$$
And this is where I get stuck!
Is this a correct way to solve this? If yes, could you help me finish this line of thought? Would Lagrange Multiplier do the job?
I really would like to solve it "the hard way" before using other properties/definitions such as:
$$|A|_2=sigma_{max}(A)$$
Thanks in advance!
linear-algebra optimization vectors norm
$endgroup$
This seems to be a very easy exercise, but as I never had previous contact with vector induced norm of a matrix, I am getting stuck...
Given the matrix:
$$A=
begin{bmatrix}
2 & 0\
0 & 1\
end{bmatrix}
$$
I am supposed to compute its vector-induced $|A|_2$ norm, defined as:
$$|A|_2=sup_{vneq 0}frac{|Av|_2}{|v|_2}$$
My attempt so far was:
Define a vector $v=(v_1,v_2)$ so that I end up with:
$$|A|_2=sup_{vneq 0}frac{|Av|_2}{|v|_2}=sup_{vneq 0}sqrt{frac{4v_{1}^{2}+v_{2}^{2}}{v_{1}^{2}+v_{2}^{2}}}$$
It seems reasonable to choose a vector $v$ such that $|v|_2=1$, then this would result in a problem where I want to find:
$$|A|_2=sup_{|v|_2=1}sqrt{4v_{1}^{2}+v_{2}^{2}}$$
Which kind of seems like an optimization problem where the constraint is:
$$sqrt{v_{1}^{2}+v_{2}^{2}}=1$$
And this is where I get stuck!
Is this a correct way to solve this? If yes, could you help me finish this line of thought? Would Lagrange Multiplier do the job?
I really would like to solve it "the hard way" before using other properties/definitions such as:
$$|A|_2=sigma_{max}(A)$$
Thanks in advance!
linear-algebra optimization vectors norm
linear-algebra optimization vectors norm
edited Jan 8 at 2:02
bertozzijr
asked Jan 8 at 1:52
bertozzijrbertozzijr
6331720
6331720
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add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Alternative:
You want to solve $$max 4v_1^2 + v_2^2$$
subject to $$v_1^2+v_2^2=1$$
Let $y_i = v_i^2$, we have
$$max 4y_1+y_2$$
subject to $$y_1+y_2=1$$
$$y_i ge 0 ,forall i in {1,2}$$
which is just a linear programming problem which has optimal solution at the corners of the polyhedral.
We want $y_1$ to be large. Hence the optimal $4y_1+y_2=4(1)+0=4$.
Taking square root gives us $2$ as the optimal solution
$endgroup$
add a comment |
$begingroup$
Here's one way to find it for a particular matrix such as the one in your example. Based on the work you've already done, observe that
$$|Av|_2^2=4v_1^2+v_2^2le4v_1^2+4v_2^2=4|v|_2^2 quad implies quad |A|_2le2.$$
Now if you can find a specific vector $v$ such that this value of $2$ is attained, i.e. such that $|Av|=2$ for a vector $v$ with $|v|_2=1$, then you will have demonstrated that $|A|_2=2$. And finding such a vector has a lot to do with eigenvectors….
$endgroup$
add a comment |
$begingroup$
Given a matrix $A$ the singular values of $A$ are just the square roots of the eigenvalues of the matrix $A^*A$. The largest singular value is the operator norm which is the quantity you want to compute.
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Alternative:
You want to solve $$max 4v_1^2 + v_2^2$$
subject to $$v_1^2+v_2^2=1$$
Let $y_i = v_i^2$, we have
$$max 4y_1+y_2$$
subject to $$y_1+y_2=1$$
$$y_i ge 0 ,forall i in {1,2}$$
which is just a linear programming problem which has optimal solution at the corners of the polyhedral.
We want $y_1$ to be large. Hence the optimal $4y_1+y_2=4(1)+0=4$.
Taking square root gives us $2$ as the optimal solution
$endgroup$
add a comment |
$begingroup$
Alternative:
You want to solve $$max 4v_1^2 + v_2^2$$
subject to $$v_1^2+v_2^2=1$$
Let $y_i = v_i^2$, we have
$$max 4y_1+y_2$$
subject to $$y_1+y_2=1$$
$$y_i ge 0 ,forall i in {1,2}$$
which is just a linear programming problem which has optimal solution at the corners of the polyhedral.
We want $y_1$ to be large. Hence the optimal $4y_1+y_2=4(1)+0=4$.
Taking square root gives us $2$ as the optimal solution
$endgroup$
add a comment |
$begingroup$
Alternative:
You want to solve $$max 4v_1^2 + v_2^2$$
subject to $$v_1^2+v_2^2=1$$
Let $y_i = v_i^2$, we have
$$max 4y_1+y_2$$
subject to $$y_1+y_2=1$$
$$y_i ge 0 ,forall i in {1,2}$$
which is just a linear programming problem which has optimal solution at the corners of the polyhedral.
We want $y_1$ to be large. Hence the optimal $4y_1+y_2=4(1)+0=4$.
Taking square root gives us $2$ as the optimal solution
$endgroup$
Alternative:
You want to solve $$max 4v_1^2 + v_2^2$$
subject to $$v_1^2+v_2^2=1$$
Let $y_i = v_i^2$, we have
$$max 4y_1+y_2$$
subject to $$y_1+y_2=1$$
$$y_i ge 0 ,forall i in {1,2}$$
which is just a linear programming problem which has optimal solution at the corners of the polyhedral.
We want $y_1$ to be large. Hence the optimal $4y_1+y_2=4(1)+0=4$.
Taking square root gives us $2$ as the optimal solution
answered Jan 8 at 2:12
Siong Thye GohSiong Thye Goh
104k1468120
104k1468120
add a comment |
add a comment |
$begingroup$
Here's one way to find it for a particular matrix such as the one in your example. Based on the work you've already done, observe that
$$|Av|_2^2=4v_1^2+v_2^2le4v_1^2+4v_2^2=4|v|_2^2 quad implies quad |A|_2le2.$$
Now if you can find a specific vector $v$ such that this value of $2$ is attained, i.e. such that $|Av|=2$ for a vector $v$ with $|v|_2=1$, then you will have demonstrated that $|A|_2=2$. And finding such a vector has a lot to do with eigenvectors….
$endgroup$
add a comment |
$begingroup$
Here's one way to find it for a particular matrix such as the one in your example. Based on the work you've already done, observe that
$$|Av|_2^2=4v_1^2+v_2^2le4v_1^2+4v_2^2=4|v|_2^2 quad implies quad |A|_2le2.$$
Now if you can find a specific vector $v$ such that this value of $2$ is attained, i.e. such that $|Av|=2$ for a vector $v$ with $|v|_2=1$, then you will have demonstrated that $|A|_2=2$. And finding such a vector has a lot to do with eigenvectors….
$endgroup$
add a comment |
$begingroup$
Here's one way to find it for a particular matrix such as the one in your example. Based on the work you've already done, observe that
$$|Av|_2^2=4v_1^2+v_2^2le4v_1^2+4v_2^2=4|v|_2^2 quad implies quad |A|_2le2.$$
Now if you can find a specific vector $v$ such that this value of $2$ is attained, i.e. such that $|Av|=2$ for a vector $v$ with $|v|_2=1$, then you will have demonstrated that $|A|_2=2$. And finding such a vector has a lot to do with eigenvectors….
$endgroup$
Here's one way to find it for a particular matrix such as the one in your example. Based on the work you've already done, observe that
$$|Av|_2^2=4v_1^2+v_2^2le4v_1^2+4v_2^2=4|v|_2^2 quad implies quad |A|_2le2.$$
Now if you can find a specific vector $v$ such that this value of $2$ is attained, i.e. such that $|Av|=2$ for a vector $v$ with $|v|_2=1$, then you will have demonstrated that $|A|_2=2$. And finding such a vector has a lot to do with eigenvectors….
answered Jan 8 at 2:04
zipirovichzipirovich
11.4k11731
11.4k11731
add a comment |
add a comment |
$begingroup$
Given a matrix $A$ the singular values of $A$ are just the square roots of the eigenvalues of the matrix $A^*A$. The largest singular value is the operator norm which is the quantity you want to compute.
$endgroup$
add a comment |
$begingroup$
Given a matrix $A$ the singular values of $A$ are just the square roots of the eigenvalues of the matrix $A^*A$. The largest singular value is the operator norm which is the quantity you want to compute.
$endgroup$
add a comment |
$begingroup$
Given a matrix $A$ the singular values of $A$ are just the square roots of the eigenvalues of the matrix $A^*A$. The largest singular value is the operator norm which is the quantity you want to compute.
$endgroup$
Given a matrix $A$ the singular values of $A$ are just the square roots of the eigenvalues of the matrix $A^*A$. The largest singular value is the operator norm which is the quantity you want to compute.
answered Jan 8 at 2:15
Mustafa SaidMustafa Said
3,0611913
3,0611913
add a comment |
add a comment |
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