Recall that $\lambda$ is an eigenvalue of the matrix $A$ if and only if the matrix $A-\lambda I$ is singular.
Thus, $A+I$ is nonsingular, otherwise $-1$ is an eigenvalue of $A$.

Since $A+I$ is nonsingular, the we have the null space $\calN(A+I)=\{\mathbf{0}\}$, and hence the nullity of $A+I$ is zero.

Given the Characteristic Polynomial, Find the Rank of the Matrix
Let $A$ be a square matrix and its characteristic polynomial is give by
\[p(t)=(t-1)^3(t-2)^2(t-3)^4(t-4).\]
Find the rank of $A$.
(The Ohio State University, Linear Algebra Final Exam Problem)
Solution.
Note that the degree of the characteristic polynomial […]

Maximize the Dimension of the Null Space of $A-aI$
Let
\[ A=\begin{bmatrix}
5 & 2 & -1 \\
2 &2 &2 \\
-1 & 2 & 5
\end{bmatrix}.\]
Pick your favorite number $a$. Find the dimension of the null space of the matrix $A-aI$, where $I$ is the $3\times 3$ identity matrix.
Your score of this problem is equal to that […]

Null Space, Nullity, Range, Rank of a Projection Linear Transformation
Let $\mathbf{u}=\begin{bmatrix}
1 \\
1 \\
0
\end{bmatrix}$ and $T:\R^3 \to \R^3$ be the linear transformation
\[T(\mathbf{x})=\proj_{\mathbf{u}}\mathbf{x}=\left(\, \frac{\mathbf{u}\cdot \mathbf{x}}{\mathbf{u}\cdot \mathbf{u}} \,\right)\mathbf{u}.\]
(a) […]

A Matrix Representation of a Linear Transformation and Related Subspaces
Let $T:\R^4 \to \R^3$ be a linear transformation defined by
\[ T\left (\, \begin{bmatrix}
x_1 \\
x_2 \\
x_3 \\
x_4
\end{bmatrix} \,\right) = \begin{bmatrix}
x_1+2x_2+3x_3-x_4 \\
3x_1+5x_2+8x_3-2x_4 \\
x_1+x_2+2x_3
\end{bmatrix}.\]
(a) Find a matrix $A$ such that […]

Orthonormal Basis of Null Space and Row Space
Let $A=\begin{bmatrix}
1 & 0 & 1 \\
0 &1 &0
\end{bmatrix}$.
(a) Find an orthonormal basis of the null space of $A$.
(b) Find the rank of $A$.
(c) Find an orthonormal basis of the row space of $A$.
(The Ohio State University, Linear Algebra Exam […]

Find Values of $h$ so that the Given Vectors are Linearly Independent
Find the value(s) of $h$ for which the following set of vectors
\[\left \{ \mathbf{v}_1=\begin{bmatrix}
1 \\
0 \\
0
\end{bmatrix}, \mathbf{v}_2\begin{bmatrix}
h \\
1 \\
-h
\end{bmatrix}, \mathbf{v}_3=\begin{bmatrix}
1 \\
2h \\
3h+1
[…]

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