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Lots and lots of matrix inequalities can be proven using the "connectedness exploit", to an extent that I am happy each time I see one that can't be solved this way.

For instance, let us call a matrix $A=\left(a_{i,j}\right)_{1\leq i\leq n,\ 1\leq j\leq n}\in\mathbb R^{n\times n}$ strictly diagonally dominant if every $i\in\left\lbrace 1,2,...,n\right\rbrace$ satisfies $a_{i,i} > \sum\limits_{j\in\left\lbrace 1,2,...,n\right\rbrace ;\ j\neq i} \left|a_{i,j}\right|$. We claim that every strictly diagonally dominant matrix $A$ has determinant $> 0$. In fact, the set of all strictly diagonally dominant matrices (as a subset of $\mathbb R^{n\times n}$) is connected (we can connect every strictly diagonally dominant matrix $A$ to a diagonal matrix with positive entries on the diagonal, just by gradually decreasing all off-diagonal entries), and every strictly diagonally dominant matrix is nonsingular (since every nonzero vector annihilated by such a matrix would lead to a contradiction, since trivial estimates would show that each of its coordinates is greater in modulus than any of its other coordinates). Qed.

Actually, we can do better: For a diagonally dominant matrix $A$ (not necessarily stricly; that is, we allow $\geq $ instead of $>$), we have

$\det A\geq \prod\limits_{i=1}^n \left(a_{i,i} - \sum\limits_{j=i+1}^n \left|a_{i,j}\right|\right)$,

where $a_{i,j}$ denote the entries of $A$. This is called Ostrowki's theorem and has been discussed elsewhere.

Another example (which used to be a Vojtech Jarnik contest problem in disguise): If $A\in\mathbb R^{n\times n}$ is a positive definite matrix and $X\in\mathbb R^{n\times n}$ is an antisymmetric matrix, then $\det\left(A+X\right)>0$. The proof is by homotopizing $X$ to $0$ while keeping it antisymmetric (all along the way, the determinant stays nonzero because $v^T\left(A+X\right)v>0$ for any $v\neq 0$). There is also an alternative proof using elementary techniques only; however it is much more complicated.

For yet another application of the same tactic, see the proof of Theorem 5.4 in this proof of van der Waerden's permanent conjecture.

Lots and lots of matrix inequalities can be proven using the "connectedness exploit", to an extent that I am happy each time I see one that can't be solved this way.

For instance, let us call a matrix $A=\left(a_{i,j}\right)_{1\leq i\leq n,\ 1\leq j\leq n}\in\mathbb R^{n\times n}$ strictly diagonally dominant if every $i\in\left\lbrace 1,2,...,n\right\rbrace$ satisfies $a_{i,i} > \sum\limits_{j\in\left\lbrace 1,2,...,n\right\rbrace ;\ j\neq i} \left|a_{i,j}\right|$. We claim that every strictly diagonally dominant matrix $A$ has determinant $> 0$. In fact, the set of all strictly diagonally dominant matrices (as a subset of $\mathbb R^{n\times n}$) is connected (we can connect every strictly diagonally dominant matrix $A$ to a diagonal matrix with positive entries on the diagonal, just by gradually decreasing all off-diagonal entries), and every strictly diagonally dominant matrix is nonsingular (since every nonzero vector annihilated by such a matrix would lead to a contradiction, because trivial estimates show that whichever of its coordinates has the greatest modulus, there must be another coordinate with yet greater modulus). Qed.

Actually, we can do better: For a diagonally dominant matrix $A$ (not necessarily strictly; that is, we allow $\geq $ instead of $>$), we have

$\det A\geq \prod\limits_{i=1}^n \left(a_{i,i} - \sum\limits_{j=i+1}^n \left|a_{i,j}\right|\right)$,

where $a_{i,j}$ denote the entries of $A$. This is called Ostrowki's theorem and has been discussed elsewhere.

Another example (which used to be a Vojtech Jarnik contest problem in disguise): If $A\in\mathbb R^{n\times n}$ is a positive definite matrix and $X\in\mathbb R^{n\times n}$ is an antisymmetric matrix, then $\det\left(A+X\right)>0$. The proof is by homotopizing $X$ to $0$ while keeping it antisymmetric (all along the way, the determinant stays nonzero because $v^T\left(A+X\right)v>0$ for any $v\neq 0$). There is also an alternative proof using elementary techniques only; however it is much more complicated.

For yet another application of the same tactic, see the proof of Theorem 5.4 in this proof of van der Waerden's permanent conjecture.

Lots and lots of matrix inequalities can be proven using the "connectedness exploit", to an extent that I am happy each time I see one that can't be solved this way.

For instance, let us call a matrix $A=\left(a_{i,j}\right)_{1\leq i\leq n,\ 1\leq j\leq n}\in\mathbb R^{n\times n}$ strictly diagonally dominant if every $i\in\left\lbrace 1,2,...,n\right\rbrace$ satisfies $a_{i,i} > \sum\limits_{j\in\left\lbrace 1,2,...,n\right\rbrace ;\ j\neq i} \left|a_{i,j}\right|$. We claim that every strictly diagonally dominant matrix $A$ has determinant $> 0$. In fact, the set of all strictly diagonally dominant matrices (as a subset of $\mathbb R^{n\times n}$) is connected (we can connect every strictly diagonally dominant matrix $A$ to a diagonal matrix with positive entries on the diagonal, just by gradually decreasing all off-diagonal entries), and every strictly diagonally dominant matrix is nonsingular (since every nonzero vector annihilated by such a matrix would lead to a contradiction, since trivial estimates would show that each of its coordinates is greater in modulus than any of its other coordinates). Qed.

Actually, we can do better: For a diagonally dominant matrix $A$ (not necessarily stricly; that is, we allow $\geq $ instead of $>$), we have

$\det A\geq \prod\limits_{i=1}^n \left(a_{i,i} - \sum\limits_{j=i+1}^n \left|a_{i,j}\right|\right)$,

where $a_{i,j}$ denote the entries of $A$. This is called Ostrowki's theorem and has been discussed elsewhere.

Another example (which used to be a Vojtech Jarnik contest problem in disguise): If $A\in\mathbb R^{n\times n}$ is a positive definite matrix and $X\in\mathbb R^{n\times n}$ is an antisymmetric matrix, then $\det\left(A+X\right)>0$. The proof is by homotopizing $X$ to $0$ while keeping it antisymmetric (all along the way, the determinant stays nonzero because $v^T\left(A+X\right)v>0$ for any $v\neq 0$). There is also an alternative proof using elementary techniques only; however it is much more complicated.

Lots and lots of matrix inequalities can be proven using the "connectedness exploit", to an extent that I am happy each time I see one that can't be solved this way.

For instance, let us call a matrix $A=\left(a_{i,j}\right)_{1\leq i\leq n,\ 1\leq j\leq n}\in\mathbb R^{n\times n}$ strictly diagonally dominant if every $i\in\left\lbrace 1,2,...,n\right\rbrace$ satisfies $a_{i,i} > \sum\limits_{j\in\left\lbrace 1,2,...,n\right\rbrace ;\ j\neq i} \left|a_{i,j}\right|$. We claim that every strictly diagonally dominant matrix $A$ has determinant $> 0$. In fact, the set of all strictly diagonally dominant matrices (as a subset of $\mathbb R^{n\times n}$) is connected (we can connect every strictly diagonally dominant matrix $A$ to a diagonal matrix with positive entries on the diagonal, just by gradually decreasing all off-diagonal entries), and every strictly diagonally dominant matrix is nonsingular (since every nonzero vector annihilated by such a matrix would lead to a contradiction, since trivial estimates would show that each of its coordinates is greater in modulus than any of its other coordinates). Qed.

Actually, we can do better: For a diagonally dominant matrix $A$ (not necessarily stricly; that is, we allow $\geq $ instead of $>$), we have

$\det A\geq \prod\limits_{i=1}^n \left(a_{i,i} - \sum\limits_{j=i+1}^n \left|a_{i,j}\right|\right)$,

where $a_{i,j}$ denote the entries of $A$. This is called Ostrowki's theorem and has been discussed elsewhere.

Another example (which used to be a Vojtech Jarnik contest problem in disguise): If $A\in\mathbb R^{n\times n}$ is a positive definite matrix and $X\in\mathbb R^{n\times n}$ is an antisymmetric matrix, then $\det\left(A+X\right)>0$. The proof is by homotopizing $X$ to $0$ while keeping it antisymmetric (all along the way, the determinant stays nonzero because $v^T\left(A+X\right)v>0$ for any $v\neq 0$). There is also an alternative proof using elementary techniques only; however it is much more complicated.

For yet another application of the same tactic, see the proof of Theorem 5.4 in this proof of van der Waerden's permanent conjecture.

Lots and lots of matrix inequalities can be proven using the "connectedness exploit", to an extent that I am happy each time I see one that can't be solved this way.

For instance, let us call a matrix $A=\left(a_{i,j}\right)_{1\leq i\leq n,\ 1\leq j\leq n}\in\mathbb R^{n\times n}$ strictly diagonally dominant if every $i\in\left\lbrace 1,2,...,n\right\rbrace$ satisfies $a_{i,i} > \sum\limits_{j\in\left\lbrace 1,2,...,n\right\rbrace ;\ j\neq i} \left|a_{i,j}\right|$. We claim that every strictly diagonally dominant matrix $A$ has determinant $> 0$. In fact, the set of all strictly diagonally dominant matrices (as a subset of $\mathbb R^{n\times n}$) is connected (we can connect every strictly diagonally dominant matrix $A$ to a diagonal matrix with positive entries on the diagonal, just by gradually decreasing all off-diagonal entries), and every strictly diagonally dominant matrix is nonsingular (since every nonzero vector annihilated by such a matrix would lead to a contradiction, since trivial estimates would show that each of its coordinates is greater in modulus than any of its other coordinates). Qed.

Actually, we can do better: For a diagonally dominant matrix $A$ (not necessarily stricly; that is, we allow $\geq $ instead of $>$), we have

$\det A\geq \prod\limits_{i=1}^n \left(a_{i,i} - \sum\limits_{j=i+1}^n \left|a_{i,j}\right|\right)$,

where $a_{i,j}$ denote the entries of $A$. This is called Ostrowki's theorem and has been discussed elsewhere.

Another example (which used to be a Vojtech Jarnik contest problem in disguise): If $A\in\mathbb R^{n\times n}$ is a positive definite matrix and $X\in\mathbb R^{n\times n}$ is an antisymmetric matrix, then $\det\left(A+X\right)>0$. The proof is by homotopizing $X$ to $0$ (all along the way, the determinant stays nonzero because $v^T\left(A+X\right)v>0$ for any $v\neq 0$). There is also an alternative proof using elementary techniques only; however it is much more complicated.

Lots and lots of matrix inequalities can be proven using the "connectedness exploit", to an extent that I am happy each time I see one that can't be solved this way.

For instance, let us call a matrix $A=\left(a_{i,j}\right)_{1\leq i\leq n,\ 1\leq j\leq n}\in\mathbb R^{n\times n}$ strictly diagonally dominant if every $i\in\left\lbrace 1,2,...,n\right\rbrace$ satisfies $a_{i,i} > \sum\limits_{j\in\left\lbrace 1,2,...,n\right\rbrace ;\ j\neq i} \left|a_{i,j}\right|$. We claim that every strictly diagonally dominant matrix $A$ has determinant $> 0$. In fact, the set of all strictly diagonally dominant matrices (as a subset of $\mathbb R^{n\times n}$) is connected (we can connect every strictly diagonally dominant matrix $A$ to a diagonal matrix with positive entries on the diagonal, just by gradually decreasing all off-diagonal entries), and every strictly diagonally dominant matrix is nonsingular (since every nonzero vector annihilated by such a matrix would lead to a contradiction, since trivial estimates would show that each of its coordinates is greater in modulus than any of its other coordinates). Qed.

Actually, we can do better: For a diagonally dominant matrix $A$ (not necessarily stricly; that is, we allow $\geq $ instead of $>$), we have

$\det A\geq \prod\limits_{i=1}^n \left(a_{i,i} - \sum\limits_{j=i+1}^n \left|a_{i,j}\right|\right)$,

where $a_{i,j}$ denote the entries of $A$. This is called Ostrowki's theorem and has been discussed elsewhere.

Another example (which used to be a Vojtech Jarnik contest problem in disguise): If $A\in\mathbb R^{n\times n}$ is a positive definite matrix and $X\in\mathbb R^{n\times n}$ is an antisymmetric matrix, then $\det\left(A+X\right)>0$. The proof is by homotopizing $X$ to $0$ while keeping it antisymmetric (all along the way, the determinant stays nonzero because $v^T\left(A+X\right)v>0$ for any $v\neq 0$). There is also an alternative proof using elementary techniques only; however it is much more complicated.