If $A$ is a $5\times5$ matrix and the dimension of the solution space of $Ax=0$ is at least $2$, then how $\operatorname{rank}(A^2) \le3, \space \det(A^2)=0$, I can't think what is the relation between $A^2$ and solution space. Thanks.
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The determinant of a square matrix of order $;n\times n;$ is zero iff it is a singular matrix iff the dimension of the solution space of the corresponding homogeneous space is greater than zero... (and many more equivalence here...). And if a square matrix is singular, then all its powers are singular matrices, too. For example, by the product theorem for determinants. – DonAntonio Nov 10 '18 at 13:53
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Since $$\operatorname{rank}(A)<5 \implies \det A=0$$ and by Binet theorem
$$\det(A^2)=\det A\cdot \det A=0$$
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@mystak That's really a foundamental fact, you need to do not forget that. You are welcome! Bye – user Nov 10 '18 at 14:01
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@gimusi, what is the meaning of 'dimension of the solution space of $Ax=0$ is at least 2.' – Win_odd Dhamnekar Nov 10 '18 at 14:30
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@DhamnekarWinod The set of all solution of $Ax=0$ is a subspace (why?) which is defined the Nullspace of A and then we can define a dimension for that in the usual way. – user Nov 10 '18 at 14:32
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Since the summation $\text{Rank}(A)$ and the dimension of the null space $N(A)$ is equal to the matrix dimension, we can conclude that $\text{Rank}(A)\leq3$. Since $\text{Rank}(AB)\leq \min\{\text{Rank}(A),\text{Rank}(B) \}$, we can easily prove $\text{Rank}(A^2)\leq3%$. Since $A^2$ is also dimension 5, this gives out the singular property.
