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Consider the following $7 \times 7$ matrix. Every unfilled entry is a $0$. What's the rank of this matrix?

$$ \begin{bmatrix} & 1 & & & & & \\ 2 & & 2 & & & & \\ & 3 & & 3 & & & \\ & & 4 & & 4 & & \\ & & & 5 & & 5 & \\ & & & & 6 & & 6 \\ & & & & & 7 & \end{bmatrix} $$

If we solve it using brute force, that is, by multiplying the scalars $c_1, c_2, \cdots, c_7$ to the column vectors and equating with $0$, we can get the rank. But is there any better way to do this? Looking at the pattern, I feel, there is.

supremacy
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1 Answers1

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Here's one way. The rank of a matrix is invariant under row and column operations (see eg. Row & Column Operation to Determine Rank). So doing the row operations $$R_3 \to R_3 - 3 R_1,$$ $$R_5 \to R_5 - \tfrac{5}{3} R_3,$$ $$R_7 \to R_7 - \tfrac{7}{5}R_5$$ in that order gets us to $$ \begin{bmatrix} & 1 & & & & & \\ 2 & & 2 & & & & \\ & & & 3 & & & \\ & & 4 & & 4 & & \\ & & & & & 5 & \\ & & & & 6 & & 6 \\ & & & & & & \end{bmatrix} $$ and then applying the column operations $$C_3 \to C_3 - C_1,$$ $$C_5 \to C_5 - C_3,$$ $$C_7 \to C_7 - C_5$$ in that order gets us to the matrix $$ \begin{bmatrix} & 1 & & & & & \\ 2 & & & & & & \\ & & & 3 & & & \\ & & 4 & & & & \\ & & & & & 5 & \\ & & & & 6 & & \\ & & & & & & \end{bmatrix} $$

Elliot Herrington
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