Eigenvalues of a Linear Transformation

[Sudharaka]

Hi everyone, :)

Here's a question I got stuck. Hope you can shed some light on it. :)

Find all eigenvalues of a linear transformation \(f\) whose matrix in some basis is \(A^{t}.A\) where \(A=(a_1,\cdots, a_n)\).

Of course if we write the matrix of the linear transformation we get,

\[A^{t}.A=\begin{pmatrix}a_1^2 & a_{1}a_2 & \cdots & a_{1}a_{n}\\a_2 a_1 & a_2^2 &\cdots & a_{2}a_{n}\\.&.&\cdots&.\\.&.&\cdots&.\\a_n a_1 & a_{n}a_2 & \cdots & a_{n}^2\end{pmatrix}\]

Now this is a symmetric matrix. So it could be written as \(A^{t}.A=QDQ^T\) where \(Q\) is a orthogonal matrix and \(D\) is a diagonal matrix. If we can do this the diagonal elements of the diagonal matrix gives all the eigenvalues we need. However I have no idea how break \(A^{t}.A\) into \(QDQ^T\). Or does any of you see a different approach to this problem which is much more easier? :)

 

[Sudharaka]

Hi everyone, :)

Here's a question I got stuck. Hope you can shed some light on it. :)
Of course if we write the matrix of the linear transformation we get,

\[A^{t}.A=\begin{pmatrix}a_1^2 & a_{1}a_2 & \cdots & a_{1}a_{n}\\a_2 a_1 & a_2^2 &\cdots & a_{2}a_{n}\\.&.&\cdots&.\\.&.&\cdots&.\\a_n a_1 & a_{n}a_2 & \cdots & a_{n}^2\end{pmatrix}\]

Now this is a symmetric matrix. So it could be written as \(A^{t}.A=QDQ^T\) where \(Q\) is a orthogonal matrix and \(D\) is a diagonal matrix. If we can do this the diagonal elements of the diagonal matrix gives all the eigenvalues we need. However I have no idea how break \(A^{t}.A\) into \(QDQ^T\). Or does any of you see a different approach to this problem which is much more easier? :)

I think I found a way to solve this problem. The method seems quite obvious but if you see any mistakes in it please let me know. :)

So we know that,

\[(A^{T}A)x=\lambda x\]

where \(x\) is the eigenvector corresponding to \(\lambda\). We simply multiply both sides by \(A\) and use the associative property of matrix multiplication.

\[A(A^{T}A)x=\lambda (Ax)\]

\[(AA^{T})(Ax)=\lambda (Ax)\]

\[(a_1^2+a^2_2+\cdots+a_n^2)(Ax)=\lambda (Ax)\]

Therefore,

\[\lambda = a_1^2+a^2_2+\cdots+a_n^2\]

And that's it! Yay, we found the eigenvalue. :p

 

[Opalg]

You have found one eigenvalue, namely $\lambda = a_1^2+a_2^2+\ldots+a_n^2$. In fact, if $x = (a_1,a_2,\ldots,a_n)^T$ then $x$ is an eigenvector, with eigenvalue $\lambda$.

Now suppose that $y = (b_1,b_2,\ldots,b_n)^T$ is a (nonzero) vector orthogonal to $x$, $x.y = 0$. If you form the product $A^TAy$, you will find that its $i$th coordinate is $a_i(x.y) = 0$ for $i=1,2,\ldots,n$, and so $A^TAy = 0$. That shows that $y$ is an eigenvector of $A^TA$, corresponding to the eigenvalue $0$. In other words, all the other eigenvalues of $A^TA$ are $0$.

 

[Sudharaka]

You have found one eigenvalue, namely $\lambda = a_1^2+a_2^2+\ldots+a_n^2$. In fact, if $x = (a_1,a_2,\ldots,a_n)^T$ then $x$ is an eigenvector, with eigenvalue $\lambda$.

Now suppose that $y = (b_1,b_2,\ldots,b_n)^T$ is a (nonzero) vector orthogonal to $x$, $x.y = 0$. If you form the product $A^TAy$, you will find that its $i$th coordinate is $a_i(x.y) = 0$ for $i=1,2,\ldots,n$, and so $A^TAy = 0$. That shows that $y$ is an eigenvector of $A^TA$, corresponding to the eigenvalue $0$. In other words, all the other eigenvalues of $A^TA$ are $0$.

Wow, thanks very much for completing my answer. It never occurred me that 0 could be a possibility of an eigenvalue. :)

 

[blue_raver22]

Hi there,

First, let's define what eigenvalues are in the context of linear transformations. Eigenvalues are the scalar values that, when multiplied by a vector, result in a new vector that is parallel to the original vector. In other words, the vector does not change direction, only its magnitude changes.

In this case, we are looking for the eigenvalues of a linear transformation whose matrix is \(A^{t}.A\). As you correctly pointed out, this is a symmetric matrix, meaning it can be diagonalized as \(A^{t}.A=QDQ^T\), where \(Q\) is an orthogonal matrix and \(D\) is a diagonal matrix.

To find the eigenvalues, we can use the fact that the eigenvalues of a diagonal matrix are simply the diagonal elements. So, to find the eigenvalues of \(A^{t}.A\), we need to find the diagonal elements of \(D\).

To do this, we can use the Singular Value Decomposition (SVD) of \(A\). SVD states that any matrix can be decomposed as \(A=UDV^T\), where \(U\) and \(V\) are orthogonal matrices and \(D\) is a diagonal matrix. In our case, we have \(A^{t}.A=(UDV^T)^{t}(UDV^T)=VD^{t}U^TUVD^TU^T=VD^{t}DV^T\).

Comparing this to our original expression of \(A^{t}.A=QDQ^T\), we can see that \(Q=V\) and \(D^{t}D=D\). Therefore, the diagonal elements of \(D\) are the eigenvalues of \(A^{t}.A\).

I hope this helps! Let me know if you have any other questions. :)