GATE Mathematics Linear Algebra Syllabus

Give an example of a matrix that is NOT diagonalizable and state why.

$A=\begin{pmatrix}1&1\\0&1\end{pmatrix}$. Its only eigenvalue is $\lambda=1$ with algebraic multiplicity $2$ but geometric multiplicity $1$ (only one independent eigenvector), so it is not diagonalizable (it is a Jordan block).

State the Spectral Theorem for real symmetric matrices.

Every real symmetric matrix $A$ ($A=A^{T}$) is orthogonally diagonalizable: there exists an orthogonal matrix $Q$ ($Q^{T}Q=I$) with $A=QDQ^{T}$, $D$ real diagonal. All eigenvalues are real and eigenvectors of distinct eigenvalues are orthogonal.

State the Cayley-Hamilton theorem.

Every square matrix satisfies its own characteristic equation: if $p(\lambda)=\det(A-\lambda I)$, then $p(A)=0$ (the zero matrix).

Define the minimal polynomial $m(\lambda)$ of a square matrix $A$.

The minimal polynomial is the unique monic polynomial $m(\lambda)$ of least degree such that $m(A)=0$.

How does the minimal polynomial relate to the characteristic polynomial?

The minimal polynomial divides the characteristic polynomial, and every eigenvalue (root of the characteristic polynomial) is also a root of the minimal polynomial. They share the same set of distinct roots.

State the diagonalizability criterion in terms of the minimal polynomial.

A matrix $A$ is diagonalizable (over a field containing its eigenvalues) if and only if its minimal polynomial factors into distinct linear factors, i.e., $m(\lambda)=(\lambda-\lambda_1)(\lambda-\lambda_2)\cdots(\lambda-\lambda_k)$ with no repeated roots.

If $A$ has characteristic polynomial $(\lambda-2)^{3}$ and minimal polynomial $(\lambda-2)^{2}$, is $A$ diagonalizable?

No. The minimal polynomial has a repeated factor $(\lambda-2)^{2}$, so $A$ is not diagonalizable.

For an eigenvalue $\lambda_i$, how does its exponent in the minimal polynomial relate to Jordan blocks?

The exponent of $(\lambda-\lambda_i)$ in the minimal polynomial equals the size of the largest Jordan block associated with $\lambda_i$.

What is the minimal polynomial of the identity matrix $I_n$ and of a scalar matrix $cI_n$?

For $I_n$: $m(\lambda)=\lambda-1$. For $cI_n$: $m(\lambda)=\lambda-c$. Both are degree $1$, reflecting that scalar matrices are already diagonal.

Why is the minimal polynomial useful for computing the inverse of $A$?

If $m(\lambda)=\lambda^{k}+c_{k-1}\lambda^{k-1}+\cdots+c_1\lambda+c_0$ with $c_0\neq0$, then from $m(A)=0$ one solves $A^{-1}=-\frac{1}{c_0}(A^{k-1}+c_{k-1}A^{k-2}+\cdots+c_1 I)$.

What is an orthonormal set of vectors?

A set $\{u_1,\dots,u_k\}$ is orthonormal if the vectors are mutually orthogonal and each has unit norm: $\langle u_i,u_j\rangle=\delta_{ij}$ (i.e., $0$ for $i\neq j$ and $1$ for $i=j$).

What is the purpose of the Gram-Schmidt orthonormalization process?

It converts a linearly independent set $\{v_1,\dots,v_n\}$ into an orthonormal set $\{u_1,\dots,u_n\}$ spanning the same subspace, producing an orthonormal basis.

Write the Gram-Schmidt formula for the orthogonal vector $w_k$ from $v_k$ given previous orthonormal vectors $u_1,\dots,u_{k-1}$.

$$w_k=v_k-\sum_{j=1}^{k-1}\langle v_k,u_j\rangle\,u_j,\qquad u_k=\frac{w_k}{\lVert w_k\rVert}.$$

In Gram-Schmidt, what is the projection of a vector $v$ onto a nonzero vector $u$?

$$\operatorname{proj}_u(v)=\frac{\langle v,u\rangle}{\langle u,u\rangle}\,u.$$

Apply the first two steps of Gram-Schmidt to $v_1=(1,1,0)$, $v_2=(1,0,1)$ in $\mathbb{R}^3$ (give $u_1$ and the orthogonal $w_2$).

$u_1=\frac{1}{\sqrt{2}}(1,1,0)$. Then $w_2=v_2-\langle v_2,u_1\rangle u_1=(1,0,1)-\frac{1}{2}(1,1,0)=\left(\frac{1}{2},-\frac{1}{2},1\right)$.

What matrix factorization is naturally produced by applying Gram-Schmidt to the columns of $A$?

The QR factorization $A=QR$, where $Q$ has orthonormal columns and $R$ is upper triangular (with the inner products / norms as entries).

Why must Gram-Schmidt be applied to a linearly independent set?

If the vectors are dependent, some $w_k$ becomes the zero vector (since $v_k$ lies in the span of previous vectors), making normalization impossible (division by zero norm).

How does the complex inner product affect the Gram-Schmidt coefficient $\langle v_k,u_j\rangle$?

In the complex case the inner product is conjugate-linear in one slot, so the coefficient $\langle v_k,u_j\rangle$ involves complex conjugation per the chosen convention; otherwise the algorithm is identical to the real case.

Define a unitary matrix $U$.

A complex square matrix $U$ is unitary if $U^{*}U=UU^{*}=I$, where $U^{*}=\overline{U}^{T}$ is the conjugate transpose. Equivalently $U^{-1}=U^{*}$; its columns form an orthonormal set under the complex inner product.

What is the real analogue of a unitary matrix, and its defining property?

An orthogonal matrix $Q$, satisfying $Q^{T}Q=QQ^{T}=I$ (i.e., $Q^{-1}=Q^{T}$); its columns are orthonormal real vectors.

Define a Hermitian matrix and state a key property of its eigenvalues.

A matrix $A$ is Hermitian if $A^{*}=A$ (equal to its conjugate transpose). All eigenvalues of a Hermitian matrix are real, and eigenvectors for distinct eigenvalues are orthogonal.

Define a normal matrix and state the condition it must satisfy.

A matrix $A$ is normal if it commutes with its conjugate transpose: $AA^{*}=A^{*}A$. Hermitian, skew-Hermitian, and unitary matrices are all normal.

State the Spectral Theorem for unitary diagonalization (which matrices are unitarily diagonalizable?).

A complex matrix $A$ is unitarily diagonalizable, i.e., $A=UDU^{*}$ with $U$ unitary and $D$ diagonal, if and only if $A$ is normal ($AA^{*}=A^{*}A$).

Write the formula for diagonalization of a normal matrix by a unitary matrix.

$$A=UDU^{*},$$ where $U$ is unitary ($U^{*}=U^{-1}$), $D=\operatorname{diag}(\lambda_1,\dots,\lambda_n)$, and the columns of $U$ are orthonormal eigenvectors of $A$.

What can be said about the eigenvalues of a unitary matrix?

All eigenvalues $\lambda$ of a unitary matrix have modulus $1$: $|\lambda|=1$, i.e., they lie on the unit circle in $\mathbb{C}$.

What are the eigenvalues of a skew-Hermitian matrix ($A^{*}=-A$)?

They are purely imaginary (of the form $i\beta$ with $\beta\in\mathbb{R}$) or zero.

How is unitary diagonalization connected to the Gram-Schmidt process?

After finding eigenvectors of a normal matrix, Gram-Schmidt is used to orthonormalize them (especially within each eigenspace for repeated eigenvalues), producing the orthonormal columns required for the unitary matrix $U$.

Compare orthogonal diagonalization and unitary diagonalization.

Orthogonal diagonalization $A=QDQ^{T}$ applies to real symmetric matrices (real $D$, orthogonal $Q$). Unitary diagonalization $A=UDU^{*}$ applies to complex normal matrices (possibly complex $D$, unitary $U$). The latter generalizes the former to the complex field.

Why is unitary/orthogonal diagonalization numerically and theoretically preferred over general $PDP^{-1}$ diagonalization?

Because $U^{-1}=U^{*}$ (no separate inverse to compute), unitary matrices preserve norms and inner products (well-conditioned, $\lVert Ux\rVert=\lVert x\rVert$), giving numerical stability and orthonormal eigenbases.

State the determinant relationship for a diagonalizable matrix $A=PDP^{-1}$ and how it gives $\det A$ and $\operatorname{trace} A$.

Determinant and trace are similarity-invariant: $\det A=\det D=\prod_i\lambda_i$ (product of eigenvalues) and $\operatorname{trace} A=\operatorname{trace} D=\sum_i\lambda_i$ (sum of eigenvalues).