Question bank — Finding eigenspaces

False — it must be singular (non-invertible), because only a singular matrix has nonzero vectors in its null space, and those are the eigenvectors.

True — $E_{\lambda }=\mathrm{Null}(A-\lambda I)$ is a subspace, and every subspace contains $0$; but $0$ is never called an eigenvector.

False — $\lambda =0$ means $A\mathbf{v}=0$ for some nonzero $\mathbf{v}$, i.e. $A$ is singular, and then $E_0=\mathrm{Null}(A)$. Only the eigenvector must be nonzero.

False — a real rotation matrix (e.g. $90^{\circ }$ rotation) turns every real vector, so no real vector stays on its line; its eigenvalues are complex.

True — $E_{\lambda }$ is a subspace (a null space), so it is closed under addition; the sum lies on the same "unrotated" set.

False — a nonzero $\mathbf{v}$ satisfying $A\mathbf{v}=3\mathbf{v}$ and $A\mathbf{v}=\mathbf{v}$ would force $3\mathbf{v}=\mathbf{v}$, impossible for $\mathbf{v}\ne 0$.

False — algebraic multiplicity is only an upper bound; $\left(\begin{array}{cc}5& 1\\ 0& 5\end{array}\right)$ repeats $\lambda =5$ twice but its eigenspace is a single line.

False — the rule is $1\le$ geometric $\le$ algebraic; the eigenspace can never be bigger than the repeat count.

True — that equality for all $\lambda$ is exactly the condition that eigenvectors form a full basis, which is what diagonalizability means.

False — if $A\mathbf{v}=\lambda \mathbf{v}$ then $A(2\mathbf{v})=\lambda (2\mathbf{v})$; scaling keeps the same $\lambda$, since eigenvalue is a property of the line, not the length.

False — the spectral theorem guarantees symmetric matrices are always diagonalizable, so geometric always equals algebraic; no deficiency occurs.

The right-hand side must be the zero vector $0$, not $\mathbf{v}$; the defining equation is $(A-\lambda I)\mathbf{v}=0$.

You subtract $\lambda$ only from the diagonal, because $A-\lambda I$ subtracts $\lambda$ times the identity, and $I$ has ones only on the diagonal.

That equation gives the eigenvalues ($\lambda$); the eigenvectors come afterward from solving the homogeneous system $(A-\lambda I)\mathbf{v}=0$ for each $\lambda$.

$0$ is never an eigenvector, and it is already in every eigenspace automatically; a spanning set must be built from nonzero independent vectors.

If the null space is just $\{0\}$ then $\lambda$ is not an eigenvalue at all; a genuine eigenspace has dimension at least $1$.

Singular means the homogeneous system has infinitely many solutions (a nonzero null space), not zero solutions; homogeneous systems always have at least $0$.

Because $E_{\lambda }$ is defined as $\mathrm{Null}(A-\lambda I)$, and finding a null space is solving $M\mathbf{v}=0$; the eigenvalue $\lambda$ just tells you which matrix $M$ to use.

$A$ is a matrix and $\lambda$ is a scalar, so $A-\lambda$ is undefined; writing $\lambda \mathbf{v}=\lambda I\mathbf{v}$ turns $\lambda$ into a genuine matrix that can be subtracted from $A$.

It is the null space of a matrix, and every null space is closed under addition and scalar multiplication and contains $0$ — the three subspace conditions.

This is the Rank-Nullity theorem: for an $n$-column matrix, rank plus nullity equals $n$, and the nullity is exactly the eigenspace dimension.

When some eigenvalue's geometric multiplicity is strictly less than its algebraic multiplicity, there aren't enough independent eigenvectors to fill a basis, so no eigenbasis exists.

$\det (A-\lambda I)=0$ is a single scalar condition on $\lambda$ that says "this matrix is singular"; it selects the special stretch factors, and only once one is chosen do vectors appear.

All of ${\mathbb{R}}^2$ — since $A-5I$ is the zero matrix, every vector satisfies $(A-5I)\mathbf{v}=0$, so no direction is special.

The only eigenvalue is $\lambda =0$ (since $A\mathbf{v}=0=0\cdot \mathbf{v}$), and $E_0={\mathbb{R}}^n$; every nonzero vector is an eigenvector.

Straight off the diagonal, because $\det (A-\lambda I)$ of a triangular matrix is the product $(a_{11}-\lambda )\cdots (a_{nn}-\lambda )$, so the roots are exactly the diagonal entries.

Each eigenvalue has algebraic multiplicity $1$, forcing geometric multiplicity $1$, so every eigenspace is a line and the matrix is automatically diagonalizable.

Yes — any scalar-multiple matrix $cI$ (like $5I$) has $E_c={\mathbb{R}}^n$, since it stretches every vector by the same factor $c$ without turning any.

It tells you $\det A=0$, because $\lambda =0$ makes $A-0\cdot I=A$ singular, and singular matrices have zero determinant.