BSc CSIT (TU) Science Mathematics II (BSc CSIT, MTH163) Question Paper 2074 Nepal

Rank of a matrix

The rank of a matrix $A$ is the number of non-zero rows in its row-echelon form, equivalently the order of the largest non-zero minor (the maximum number of linearly independent rows or columns). It is denoted $\rho(A)$ or $\operatorname{rank}(A)$.

Rank by reduction to echelon form

Consider

Apply elementary row operations:

• $R_2\to R_2-2R_1,\quad R_3\to R_3-3R_1$:

• $R_3\to R_3-R_2$:

There are 2 non-zero rows, so $\rho(A)=2$.

Solving a system by Gauss elimination

Solve

Form the augmented matrix and eliminate:

$R_3\to 3R_3+R_2$:

Back-substitution:

• $-7z=-21\Rightarrow z=3$.
• $-3y-z=-9\Rightarrow -3y-3=-9\Rightarrow y=2$.
• $x+y+z=6\Rightarrow x=6-2-3=1$.

Solution: $x=1,\ y=2,\ z=3$.

Cayley–Hamilton theorem (statement)

Every square matrix satisfies its own characteristic equation. If $A$ is an $n\times n$ matrix with characteristic polynomial $p(\lambda)=\det(A-\lambda I)$, then $p(A)=O$ (the zero matrix).

Proof (outline)

Let $p(\lambda)=\det(A-\lambda I)$. Consider the adjoint $\operatorname{adj}(A-\lambda I)$, whose entries are polynomials in $\lambda$ of degree at most $n-1$. By the adjoint property,

Write $\operatorname{adj}(A-\lambda I)=B_{n-1}\lambda^{n-1}+\dots+B_1\lambda+B_0$ with matrix coefficients $B_i$, and $p(\lambda)=\sum_{k=0}^{n}c_k\lambda^{k}$. Equating like powers of $\lambda$ on both sides of $(A-\lambda I)\,\operatorname{adj}(A-\lambda I)=p(\lambda)I$ gives relations among the $B_i$. Multiplying these matrix equations successively by $I,A,A^2,\dots,A^n$ and adding causes all terms to telescope, yielding $\sum_k c_k A^{k}=p(A)=O$. $\blacksquare$

Finding $A^{-1}$ using the theorem

Let

Characteristic polynomial (with $\operatorname{tr}A=-2$, sum of principal $2\times2$ minors $=-6$, $\det A=6$):

So $A^{3}+2A^{2}-6A-6I=O$.

Multiply by $A^{-1}$:

Compute $A^{2}$, substitute, and the result is $A^{-1}=\tfrac{1}{6}(A^{2}+2A-6I)$, which can be evaluated entrywise to obtain the inverse.

Method: the Cayley–Hamilton relation lets us express $A^{-1}$ as a polynomial in $A$, avoiding the adjoint/cofactor computation.

Definitions

For a square matrix $A$, a scalar $\lambda$ is an eigenvalue if there exists a non-zero vector $X$ with

The vector $X$ is the corresponding eigenvector. Eigenvalues are the roots of the characteristic equation $\det(A-\lambda I)=0$.

Worked example

Let

Characteristic equation:

So $(2-\lambda)=0$ or $(2-\lambda)^2=1$, giving eigenvalues

Eigenvectors:

• $\lambda=1$: $(A-I)X=0\Rightarrow x_1+x_3=0,\ x_2=0\Rightarrow X_1=(1,0,-1)^T$.
• $\lambda=2$: $(A-2I)X=0\Rightarrow x_3=0,\ x_1=0,\ x_2$ free $\Rightarrow X_2=(0,1,0)^T$.
• $\lambda=3$: $(A-3I)X=0\Rightarrow -x_1+x_3=0,\ x_2=0\Rightarrow X_3=(1,0,1)^T$.

Eigenvalues $1,2,3$ with eigenvectors $(1,0,-1)^T,\,(0,1,0)^T,\,(1,0,1)^T$ respectively.

Symmetric matrix: a square matrix $A$ that equals its transpose, $A^{T}=A$ (so $a_{ij}=a_{ji}$). Example: $\begin{pmatrix}1&2\\2&3\end{pmatrix}$.

Orthogonal matrix: a square matrix $A$ whose transpose is its inverse, $A^{T}A=AA^{T}=I$, equivalently $A^{-1}=A^{T}$. Its rows (and columns) form an orthonormal set and $\det A=\pm1$. Example: $\begin{pmatrix}\cos\theta&-\sin\theta\\\sin\theta&\cos\theta\end{pmatrix}$, since $A^{T}A=I$.

Reduce $A=\begin{pmatrix}1&2&3\\2&4&6\\1&1&1\end{pmatrix}$ to echelon form.

• $R_2\to R_2-2R_1$, $R_3\to R_3-R_1$:

• Swap $R_2\leftrightarrow R_3$:

There are 2 non-zero rows, so $\rho(A)=2$. (Note row 2 is twice row 1, so the rank cannot be 3.)

Properties of determinants:

1. $\det(A^{T})=\det(A)$ — value is unchanged by transposing.
2. Interchanging two rows (or columns) multiplies the determinant by $-1$.
3. If any two rows (or columns) are identical or proportional, $\det=0$.
4. Multiplying one row/column by a scalar $k$ multiplies the determinant by $k$.
5. Adding a multiple of one row (column) to another leaves the determinant unchanged.
6. If a row or column is all zeros, $\det=0$.
7. $\det(AB)=\det(A)\det(B)$; for an $n\times n$ matrix, $\det(kA)=k^{n}\det(A)$.
8. For a triangular (or diagonal) matrix, $\det$ equals the product of the diagonal entries; $\det(I)=1$. Also $\det(A^{-1})=1/\det(A)$.

The null space (or kernel) of an $m\times n$ matrix $A$ is the set of all vectors $X$ satisfying $AX=0$:

It is a subspace of $\mathbb{R}^{n}$ (it contains $0$ and is closed under addition and scalar multiplication). Its dimension is the nullity of $A$, and by the rank–nullity theorem $\operatorname{rank}(A)+\text{nullity}(A)=n$.

Example: for $A=\begin{pmatrix}1&1\\2&2\end{pmatrix}$, $AX=0$ gives $x+y=0$, so $N(A)=\{t(1,-1)^T\}$, a 1-dimensional subspace.

A basis of a vector space $V$ is a set of vectors $\{v_1,v_2,\dots,v_n\}$ that is

1. linearly independent, and
2. spans $V$ (every vector in $V$ is a linear combination of them).

Thus a basis is a minimal spanning set / maximal linearly independent set. The number of vectors in any basis is the dimension of $V$, and every vector has a unique representation in a given basis.

Example: $\{(1,0),(0,1)\}$ is the standard basis of $\mathbb{R}^{2}$.

For a diagonal matrix the eigenvalues are simply the diagonal entries. Formally:

Hence the eigenvalues are $\lambda_1=2$ and $\lambda_2=3$ (with eigenvectors $(1,0)^T$ and $(0,1)^T$).

Consider the linear combination set to zero:

This forces $c_1=0$ and $c_2=0$. Since the only solution is the trivial one, the vectors $(1,0)$ and $(0,1)$ are linearly independent.

Equivalently, the matrix $\begin{pmatrix}1&0\\0&1\end{pmatrix}$ has $\det=1\neq0$, confirming independence.

The adjoint (adjugate) of a square matrix $A$ is the transpose of its cofactor matrix:

where $M_{ij}$ is the minor obtained by deleting row $i$ and column $j$. It satisfies the key identity

so for $\det(A)\neq0$, $A^{-1}=\dfrac{1}{\det(A)}\operatorname{adj}(A)$.

Example: for $A=\begin{pmatrix}a&b\\c&d\end{pmatrix}$, $\operatorname{adj}(A)=\begin{pmatrix}d&-b\\-c&a\end{pmatrix}$.

A singular matrix is a square matrix whose determinant is zero, $\det(A)=0$. Consequences:

• It has no inverse (it is non-invertible).
• Its rows/columns are linearly dependent, so its rank is less than its order.
• The system $AX=0$ has non-trivial solutions, and $\lambda=0$ is an eigenvalue.

Example: $\begin{pmatrix}1&2\\2&4\end{pmatrix}$ has $\det=1\cdot4-2\cdot2=0$, so it is singular. (A matrix with $\det\neq0$ is called non-singular.)