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I would like to know if $A$ is a singular square matrix. Then by adding a scaled diagonal term $\lambda I$ will it become non-singular? Would it matter if the matrix $A$ is symmetric or not?

If it does make it non-singular how can I prove that?

Edit:

What if I add the following constraints: A is positive semi-definite and $\lambda$ is strictly positive?

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    It **can** provided $\lambda\ne -\mu$ for any eigenvalue, $\mu$, of $A$. – 2017-02-15
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    Isn't the solution to $\det(A-\lambda{I})=0$ exactly the values of $\lambda$ for which $A-\lambda{I}$ is singular? – 2017-02-15
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    I didn't mean for $\lambda$ to be an eigenvalue, my question was not so clear sorry about that. – 2017-02-15
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    With your constraints, $A + \lambda I$ will be positive definite and therefore non-singular. Use the definition of positive definite. – 2017-02-15
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    right! $v^TAv \geq 0$ and $v^T(\lambda I)v > 0$ hence their sum is positive definite. Thanks! – 2017-02-15

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Consider $$A=\begin{pmatrix}-1&0\\0&0\end{pmatrix}$$

Then $A$ is singular and $A+I$ is singular.

In general, if $A$ is singular and has some nonzero eigenvalue, the statement is false.

EDIT: Now, if $A$ is positive semidefinite, then every eigenvalue is nonnegative. Then $A+\lambda I$ is singular implies that $-\lambda$ is an eigenvalue and hence $\lambda\le 0$. This contradicts the other constraint. So yes, with these constraints, the statement is true (and note that you don't need the singularity of $A$ anymore; actually, the singularity of $A$ and of $A+\lambda I$ have little to do each other).

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    Thanks ajotatxe can you check my edit? – 2017-02-15
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$A-\lambda$ is singular when $\lambda$ is an eigenvalue. Hence, if you choose a different value for $\lambda$, it will be non-singular.

For example, $A=\left(\begin{matrix}0 & 0\\0 & 1\end{matrix}\right)$ is a singular matrix with eigenvalues $0$ and $1$, and so $A=A-0$ and $A-1$ are both singular. $A-2$ is non-singular.

The story is the same if $A$ is not singular. For example, $A=\left(\begin{matrix}1 & 0\\0 & 2\end{matrix}\right)$ is non-singular, $A-1$ and $A-2$ are singular, and $A-3$ is non-singular.

You may find the eigenvalues of a matrix by solving the equation $\det(A-\lambda)=0$ for $\lambda$.

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Using the constraints, and following Omnomnomnom suggestion to use the definition of positive definite:

$$v^T(A + \lambda I) v = v^TAv + v^T\lambda I v$$

Since $v^TAv \geq 0$ and $v^T\lambda I v > 0$, then their sum is positive definite which is non-singular.