TIFR CSE 2021 | Part A | Question: 3

Question

Let $M$ be an $n \times m$ real matrix. Consider the following:

• Let $k_{1}$ be the smallest number such that $M$ can be factorized as $A \cdot B$, where $A$ is an $n \times k_{1}$ and $B$ is a $k_{1} \times m$ matrix.
• Let $k_{2}$ be the smallest number such that $M=\sum_{i=1}^{k_{2}}u_{i}v_{i},$ where each $u_{i}$ is an $n \times 1$ matrix and each $v_{i}$ is an $1 \times m$ matrix.
• Let $k_{3}$ be the column-rank of $M$.

Which of the following statements is $\text{TRUE}$?

• A. $k_{1}< k_{2}< k_{3}$
• B. $k_{1}< k_{3}< k_{2}$
• C. $k_{2}= k_{3}< k_{1}$
• D. $k_{1}= k_{2}= k_{3}$
• E. No general relationship exists among $k_{1}, k_{2}$ and $k_{3}$

Answer 1

first method and second method are basically same things. They are just alternative ways to look at matrix multiplication.

Matrix multiplication can be treated as taking linear combinations of first matrix with scalars as columns of second matrix. or another way to look at it is taking linear combinations of second matrix with scalars as rows of first matrix, Consider either way of matrix multiplication in this example.

Rank represents number of independent columns/rows present in matrix.

for a 3*3 matrix M if rank is 3 the same matrix can be given as M*I.

for rank<3, suppose 2, columns can be treated as c1,c2= x*c1, c3.(or r1, x*r1, r3).

now putting c1 and c3 in one matrix and multiplying them with appropriate matrix we can get original matrix as c2(x*c1) is scalar multiple of c1. And c1,c3 can be derived in this multiplication from themselves easily. similar is case if we consider the row way. just the way of multiplication will be changed(i.e. first matrix will be holding multiples and second will be holding independent rows of original matrix).

hence rank can be used to decompose matrix in two matrices.

From all of this we can conclude that k1=k2=k3.

Comments

how rank can be used to decompose matrices??