MHB 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. :)