Show that the matrix $A^2 + I$ is invertible for all matrices $A$, where $A$ is an $n \times n$ symmetric matrix. I'm a little stuck on this problem. I know that since $A$ is symmetric, $A=A^{T}$. I'm also pretty sure that $AA^{T}$ is invertible. Therefore $A^2$ would be invertible. I'm not really sure how to account for the identity matrix (In).
Thanks in advance!
 A: For an arbitrary real valued matrix $A$, $A^TA$ is positive-semidefinite, so all its eigenvalues are nonnegative. It follows that all the eigenvalues of $A^TA+I$ are no less than $1$, so $A^TA+I$ is invertible(and positive definite). In particular, when $A$ is symmetric, $A^2+I$ is invertible(and positive definite). 
A more direct way is to show that for any nonzero column vector $v$, $(A^TA+I)v\ne 0$. To see this, note that for every column vector $w$, $w^Tw\ge 0$ and the equality holds if and only if $w=0$. Then 
$$v^T(A^TA+I)v=(Av)^T(Av)+v^Tv>0.\tag{1}$$
$(1)$ implies that $(A^TA+I)v\ne 0$, which completes the proof. In fact, $(1)$ also shows that $A^TA+I$ is positive definite.
A: Hint: Since $A$ is a symmetric matrix, its eigen value are real ( it can be zero as well ) however when the identity matrix is added, the eigen value will become non-zero real values.
A: Problem
Show that the matrix A2 + In is invertible for all matrices A, where A is an n×n symmetric matrix.
Solution
Show that in another basis A2 + In is a diagonal matrix with no zeros in the diagonal — in other words, it only scales up or reflects as a transformation, and is thus invertible. 
In Great Detail
The spectral theorem states that if  A is symmetric, there exists an invertible C matrix and a real diagonal matrix B such that A = CBC-1. A diagonal matrix only has values or zeros in the diagonal. (More info: Wikipedia) That conversion is just between bases. (More info: Khan Academy)
So if we substitute in that CBC-1 for A...
A2 + In = (CBC-1)2 + In
 = CBC-1CBC-1 + In
But we also know that In = C-1C = CC-1, so we can change that last bit up a bit...
CBC-1CBC-1 + In = CB In BC-1 + CC-1 = CBBC-1 + CC-1 = CB2C-1 + CC-1
 Matrix multiplication isn't commutative (xy=yx) but it is associative ((xy)z=x(yz)) and distributive (x(y+z)=xy+xz, (y+z)x=yx+zx), so we can change that into another form. 
CB2C-1 + CC-1 = C(B2C-1 + C-1) = C((B2 + In)C-1) = C(B2 + In)C-1
Why is this important? Well, don't forget that B was a diagonal matrix!  B2 is going to have positive numbers down the diagonal or zeroes. Since we add the identity matrix In, we know that every diagonal entry will have a value of at least 1 (a non-zero value). 
So, this shows that A2 + In can be expressed as a diagonal matrix of the form B2 + In in another basis where all the diagonal entries are equal to or greater than 1. 
What does invertible mean? It means you can undo a transformation by multiplying by another matrix. B2 + In is a diagonal matrix with no zeros, so each component of a vector mutliplied by B2 + In is going to be scaled up in some way or another. In other words, the transformed vector will be only scaled up/down or reflected — in other words, B2 + In is a transformation that can be undone. It's invertible. 
Well, that doesn't tell us about A2 + In, you might say. But wait! B2 + In is just another form of A2 + In in another basis. You can convert transformations between bases but still have it be the same transformation. Thus, (B2 + In)'s inverse can be converted back into (A2 + In)'s coordinates. That's the same transformation but in different bases. (B2 + In)'s inverse is (A2 + In)'s inverse.
Thus, we conclude that A2 + In has an inverse. It is therefore invertible. 
A: Eigenvalues of symmetric matrix are purely real, Enough to show $\det(A^2+I)\ne 0$ i.e enough to show that $-1$ is not an eigen value of $A^2$, $A$ symmetric so if $\lambda_i$'s are eigen values  then $\lambda_i\in\mathbb{R}\forall i$ and the eigen values of $A^2$ will be $\lambda_i^2\ge 0$ so indeed $-1$ can not be an eigen value of $A^2$ 
