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Let $A=[a_{i, j}]$ be an $n\times n$ matrix with real entries. spz that there is an $m$ with $a_{i, j} =0$ for $i\ge m$, $j\le m$ and $a_{i, i}$ not equal to zero for $1\le i<m$. Prove that $A$ is singular.

I understand that a matrix is non singular when the either the det not equal to zero or the linear combination is not equal to zero making the column vectors linearly dependent. Although, I am having trouble (1) seeing hat this matrix written out would look like and (2)proving it is singular.

Lutz Lehmann
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Sherien Hassan
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1 Answers1

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Your matrix has the block form $$ A=\pmatrix{A_{11}&A_{12}\\0~..~0&a_{m,m+1..n}\\0&A_{32}} $$ where the second row of the block matrix is the $m$th row of the original matrix.

Looking at the first $m$ columns, they have to be linearly dependent, as the non-zero block $A_{11}$ is an $(m-1)\times m$ matrix, consequently with rank at most $m-1$. The full matrix is thus also singular.

Lutz Lehmann
  • 126,666
  • I am sorry I am unsure why this is hard for me to see. So for example would the matrix (2 6 1 6) (0 0 1 4) ( 0 0 4 1) (0 0 6 2) be an example? – Sherien Hassan Oct 27 '19 at 18:17
  • Yes, that matrix fits the description. – Lutz Lehmann Oct 27 '19 at 18:43
  • So then, i would bee to multiply this matrix by a list of coefficient, and then write out the linear combination and set it equal to zero. But I do not see where what part of this proves it is singular? Or how to show it for any size matrix. I know that in terms of singularity of matrices a whole row of column because equal to zero would cause for it to be singular and here no whole row or whole column is. – Sherien Hassan Oct 27 '19 at 22:23
  • You need that all columns are linearly independent. But the first $m$ columns are already dependent, there exists a non-trivial linear combination of these that gives zero. – Lutz Lehmann Oct 28 '19 at 06:31