# [SOLVED]Eigenvalues of a Linear Transformation

#### Sudharaka

##### Well-known member
MHB Math Helper
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

##### Well-known member
MHB Math Helper
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. #### Opalg

##### MHB Oldtimer
Staff member
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

##### Well-known member
MHB Math Helper
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. 