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Let's say we have a matrix with 31 rows whose rank is 30. In addition if we remove any one row from the matrix, so it has 30 rows, then the rank will remain 30, no matter which row is removed.

Does this mean that every row in the matrix is equal to some linear combination of the other 30 rows?

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  • $\begingroup$ Yes. ${}{}{}{}{}$ $\endgroup$
    – copper.hat
    Jan 5, 2018 at 6:29

2 Answers 2

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Call the first 30 rows $R_1,R_2,\ldots, R_{30}$. T

Choose 30 numbers $a_1,a_2,\ldots, a_{30}$ each of them non-zero. Define $R_{31}$ to be $a_1R_1+a_2R_2+\cdots +a_{30}R_{30}$. Putting this as the 31st row gives the situation you describe.

As $a_7$ is non-zero (for example) we can express $R_7$ as as linear combination the remaining 30 rows from $R_1$ to $R_{31}$. This is the only way you can get this.

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Yes. If removing a row does not change the dimension of the row space that row is a linear combination of the other rows. In your case any row is a linear combination of the other $30$ rows.

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