Eigenvalue problem

[marc146]

Ok so I have two systems
A=[[0,1,2],[3,-2,1],[2,-4,-5]]
and
b= [[2],[0],[1]].
The goal is to determine the column vector g so that the eigevalues of A_c= A-b*g^(T) are at -3,-2,2+- i*3*sqrt(3/2).

The textbook for class is advanced engineering by Greenberg. But there is nothing remotely close to solving this problem in there. Any help would be appreciated.

 

[galactus]

Ok so I have two systems
A=[[0,1,2],[3,-2,1],[2,-4,-5]]
and
b= [[2],[0],[1]].
The goal is to determine the column vector g so that the eigevalues of A_c= A-b*g^(T) are at -3,-2,2+- i*3*sqrt(3/2).

The textbook for class is advanced engineering by Greenberg. But there is nothing remotely close to solving this problem in there. Any help would be appreciated.

Same as here:

\(\displaystyle A=\begin{bmatrix}0 & 1 & -2\\3 & -2 & 1 \\2 & -4 & -5\end{bmatrix}\)
b=\(\displaystyle \begin{bmatrix}2 \\ 0 \\1\end{bmatrix}\)

and i have to determine the column vector g so that the eigenvalues of \(\displaystyle A{c}=A-b\cdot g^{T}\) (g transpose) are at \(\displaystyle -3, -2\pm3i\sqrt{\frac{3}{2}}\)

To clarify, what does the 'c' stand for?. I assume a column matrix.

The characteristic polynomial would be \(\displaystyle x^{3}+7x^{2}+\frac{59}{2}x+\frac{105}{2}=0\)

Putting it together gives:

\(\displaystyle \begin{bmatrix}0&1&-2\\3&-2&1\\2&-4&-5\end{bmatrix}\cdot\begin{bmatrix}c_{1}\\c_{2}\\c_{3}\end{bmatrix}=\begin{bmatrix}0&1&-2\\3&-2&1\\2&-4&-5\end{bmatrix}-\begin{bmatrix}2\\0\\1\end{bmatrix}\cdot \begin{bmatrix}g_{1}&g_{2}&g_{3}\end{bmatrix}\)

Multiplying and simplifying gives:

\(\displaystyle \begin{bmatrix}0&c_{2}&-2c_{3}\\3c_{1}&-2c_{2}&c_{3}\\2c_{1}&-4c_{2}&-5c_{3}\end{bmatrix}=\begin{bmatrix}-2g_{1}&1-2g_{2}&-2g_{3}-2\\3&-2&1\\2-g_{1}&-g_{2}-4&-g_{3}-5\end{bmatrix}\)

Can you get any ideas from here?. I will look a little more tomorrow.

 

[marc146]

Nice making it look neat.

It is A sub c. A_(c) is the latex code I believe. I have no idea what it means other than how it was written.

 

[marc146]

Ok so I have two systems
A=[[0,1,2],[3,-2,1],[2,-4,-5]]
and
b= [[2],[0],[1]].
The goal is to determine the column vector g so that the eigevalues of A_c= A-b*g^(T) are at -3,-2,2+- i*3*sqrt(3/2).

The textbook for class is advanced engineering by Greenberg. But there is nothing remotely close to solving this problem in there. Any help would be appreciated.

Same as here:

\(\displaystyle A=\begin{bmatrix}0 & 1 & -2\\3 & -2 & 1 \\2 & -4 & -5\end{bmatrix}\)
b=\(\displaystyle \begin{bmatrix}2 \\ 0 \\1\end{bmatrix}\)

and i have to determine the column vector g so that the eigenvalues of \(\displaystyle A{c}=A-b\cdot g^{T}\) (g transpose) are at \(\displaystyle -3, -2\pm3i\sqrt{\frac{3}{2}}\)

To clarify, what does the 'c' stand for?.

\(\displaystyle \begin{bmatrix}0&1&-2\\3&-2&1\\2&-4&-5\end{bmatrix}\cdot\begin{bmatrix}c_{1}\\c_{2}\\c_{3}\end{bmatrix}=\begin{bmatrix}0&1&-2\\3&-2&1\\2&-4&-5\end{bmatrix}-\begin{bmatrix}2\\0\\1\end{bmatrix}\cdot \begin{bmatrix}g_{1}&g_{2}&g_{3}\end{bmatrix}\)

This gives:

\(\displaystyle \begin{bmatrix}0&c_{2}&-2c_{3}\\3c_{1}&-2c_{2}&c_{3}\\2c_{1}&-4c_{2}&-5c_{3}\end{bmatrix}=\begin{bmatrix}-2g_{1}&1-2g_{2}&-2g_{3}-2\\3&-2&1\\2-g_{1}&-g_{2}-4&-g_{3}-5\end{bmatrix}\)

Ok, I have something similar:

I have the matrix A-b* \(\displaystyle \begin{bmatrix}g_{1}&g_{2}&g_{3}\end{bmatrix}\)

I multiplied b*g and subtracted from A giving ONLY
\(\displaystyle \begin{bmatrix}-2g_{1}&1-2g_{2}&-2-2g_{3}\\3&-2&1\\2-g_{1}&-4-g_{2}&-5-g_{3}\end{bmatrix}\)

So, I am left which ONE matrix, but you have two? Clearly you are the expert, but why are you setting them equal to each other?
Also, how does your intent to set them equal each other bring about the overall concept of getting those eigenvalues?

 

[galactus]

I am no expert. Linear algebra is not my forte, but I know enough to get into trouble

I set them equal because that is what the problem does. \(\displaystyle A_{c}=A-bg^{T}\)

The thing is, I do not know what the subscript 'c' is on the left side. I just assumed, probably incorrectly, it was a column vector.

 

[marc146]

Ok, I can see why you did that....based on the A_{c} given in the problem. Ok, so you think I should just proceed with the characteristic polynomial on the RHS then?

Again, I have no idea what that subscript 'c' stands for, it is just what the problem stated. That is why I simply ignored it when I first started.

 

[marc146]

Ok so I have two systems
A=[[0,1,2],[3,-2,1],[2,-4,-5]]
and
b= [[2],[0],[1]].
The goal is to determine the column vector g so that the eigevalues of A_c= A-b*g^(T) are at -3,-2,2+- i*3*sqrt(3/2).

The textbook for class is advanced engineering by Greenberg. But there is nothing remotely close to solving this problem in there. Any help would be appreciated.

Same as here:

\(\displaystyle A=\begin{bmatrix}0 & 1 & -2\\3 & -2 & 1 \\2 & -4 & -5\end{bmatrix}\)
b=\(\displaystyle \begin{bmatrix}2 \\ 0 \\1\end{bmatrix}\)

and i have to determine the column vector g so that the eigenvalues of \(\displaystyle A{c}=A-b\cdot g^{T}\) (g transpose) are at \(\displaystyle -3, -2\pm3i\sqrt{\frac{3}{2}}\)

To clarify, what does the 'c' stand for?. I assume a column matrix.

The characteristic polynomial would be \(\displaystyle x^{3}+7x^{2}+\frac{59}{2}x+\frac{105}{2}=0\)

Putting it together gives:

\(\displaystyle \begin{bmatrix}0&1&-2\\3&-2&1\\2&-4&-5\end{bmatrix}\cdot\begin{bmatrix}c_{1}\\c_{2}\\c_{3}\end{bmatrix}=\begin{bmatrix}0&1&-2\\3&-2&1\\2&-4&-5\end{bmatrix}-\begin{bmatrix}2\\0\\1\end{bmatrix}\cdot \begin{bmatrix}g_{1}&g_{2}&g_{3}\end{bmatrix}\)

Multiplying and simplifying gives:

\(\displaystyle \begin{bmatrix}0&c_{2}&-2c_{3}\\3c_{1}&-2c_{2}&c_{3}\\2c_{1}&-4c_{2}&-5c_{3}\end{bmatrix}=\begin{bmatrix}-2g_{1}&1-2g_{2}&-2g_{3}-2\\3&-2&1\\2-g_{1}&-g_{2}-4&-g_{3}-5\end{bmatrix}\)

Can you get any ideas from here?. I will look a little more tomorrow.

Don't know how you came up with that characteristic polynomial. My idea (that I'm going to plug into Maple) is to find the char polynomial of \(\displaystyle \begin{bmatrix}-2g_{1}&1-2g_{2}&-2g_{3}-2\\3&-2&1\\2-g_{1}&-g_{2}-4&-g_{3}-5\end{bmatrix}\) and then somehow setting them equal to the eigenvalues given in the problem and solve for g1, g2, g3. Don't know how feasible that is, but going to work it.

 

[marc146]

Ok so I have two systems
A=[[0,1,2],[3,-2,1],[2,-4,-5]]
and
b= [[2],[0],[1]].
The goal is to determine the column vector g so that the eigevalues of A_c= A-b*g^(T) are at -3,-2,2+- i*3*sqrt(3/2).

The textbook for class is advanced engineering by Greenberg. But there is nothing remotely close to solving this problem in there. Any help would be appreciated.

Same as here:

\(\displaystyle A=\begin{bmatrix}0 & 1 & -2\\3 & -2 & 1 \\2 & -4 & -5\end{bmatrix}\)
b=\(\displaystyle \begin{bmatrix}2 \\ 0 \\1\end{bmatrix}\)

and i have to determine the column vector g so that the eigenvalues of \(\displaystyle A{c}=A-b\cdot g^{T}\) (g transpose) are at \(\displaystyle -3, -2\pm3i\sqrt{\frac{3}{2}}\)

To clarify, what does the 'c' stand for?. I assume a column matrix.

The characteristic polynomial would be \(\displaystyle x^{3}+7x^{2}+\frac{59}{2}x+\frac{105}{2}=0\)

Putting it together gives:

\(\displaystyle \begin{bmatrix}0&1&-2\\3&-2&1\\2&-4&-5\end{bmatrix}\cdot\begin{bmatrix}c_{1}\\c_{2}\\c_{3}\end{bmatrix}=\begin{bmatrix}0&1&-2\\3&-2&1\\2&-4&-5\end{bmatrix}-\begin{bmatrix}2\\0\\1\end{bmatrix}\cdot \begin{bmatrix}g_{1}&g_{2}&g_{3}\end{bmatrix}\)

Multiplying and simplifying gives:

\(\displaystyle \begin{bmatrix}0&c_{2}&-2c_{3}\\3c_{1}&-2c_{2}&c_{3}\\2c_{1}&-4c_{2}&-5c_{3}\end{bmatrix}=\begin{bmatrix}-2g_{1}&1-2g_{2}&-2g_{3}-2\\3&-2&1\\2-g_{1}&-g_{2}-4&-g_{3}-5\end{bmatrix}\)

Can you get any ideas from here?. I will look a little more tomorrow.

I tried a radical idea using Maple that I stated at the bottom. It doesn't work out. What do you think?

 

[galactus]

The char poly is found from \(\displaystyle Det(A-{\lambda}I)\)

Letting A be what we came up with in terms of \(\displaystyle g_{k}, \;\ k=1,2,3\), the charpoly is

\(\displaystyle -2g_{1}{\lambda}^{2}-12g_{1}{\lambda}-25g_{1}-7g_{2}{\lambda}-28g_{2}-g_{3}{\lambda}^{2}-6g_{3}{\lambda}+19g_{3}-{\lambda}^{3}-7{\lambda}^{2}-15{\lambda}+33=0\)

\(\displaystyle (-2{\lambda}^{2}-12{\lambda}-25)g_{1}-(7{\lambda}-28)g_{2}-({\lambda}^{2}+6{\lambda}-19)g_{3}-({\lambda}^{3}+7{\lambda}^{2}+15)+33=0\)

Setting lambda equal to the three given roots gives three equations with three unknowns, \(\displaystyle g_{1}, g_{2}, g_{3}\)

\(\displaystyle -7g_{1}-7g_{2}+18g_{3}+42=0\)

\(\displaystyle (18-6\sqrt{6}i)g_{1}-(\frac{21\sqrt{6}}{2}i+14)\cdot g_{2}+(\frac{81}{2}-3\sqrt{6}i)\cdot g_{3}+\frac{113}{2}+\frac{87\sqrt{6}}{4}i=0\)

\(\displaystyle (18+6\sqrt{6}i)g_{1}-(14-\frac{21\sqrt{6}}{2}i)g_{2}+(\frac{81}{2}+3\sqrt{6}i)g_{3}+\frac{113}{2}-\frac{87\sqrt{6}}{4}i=0\)

Now, solve for the respective g's.

See if that works. Make sure my equations are correct. One sign wrong and it is all wrong.