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I think it is true because I can't find a matrix that isin't following the above statement. A little help would be appreciated!

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    What is "equivalent" for you? – 2017-01-19
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    We can apply row and column elementary transformations on the original matrix. – 2017-01-19
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    Then yes, it is true. Hint: at least one row is a linear combination of other rows. – 2017-01-19

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Two matrices $A,B$ are equivalent, when there are two invertiable matrices $U,V$ so that $B=UAV$ holds. If we use a singular value decomposition $A = U \Sigma V^T$, you get $\Sigma$ as diagonal matrix with the singular values of $A$ on the diagonal. Since the determinant of $A$ is zero, there is at least one singular value $\sigma_n=0$. Since $U$ and $V$ are orthogonal matrices, they are invertible.

Short: $A$ is equivalent to $\Sigma$, the matrix of it's singular values. One of the singular values is zero, therefore a complete collumn and a complete row is zero.

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    Singular value decomposition? The problem is much easier than that. – 2017-01-19
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    But it makes an easy proof. Just because you can use more basic techniques, does not mean, you shouldn't do it... sophisticated. I'm also deeply routed in singular values, eigenvalues so that's what first comes to my mind, especially when asked about equivlant or similar matrices. – 2017-01-19
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    Singular values are completely irrelevant here. What does matter is rank. – 2017-01-19
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    And since the number of singular values is the rank of the matrix, we are using the same argument. I just showed an easy "standard" way of obtaining such a matrix. – 2017-01-19