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11

We will use the following Lemmas: ${\bf Lemma~1.}$ Let $R$, $S$ be two real $2\times 2$ matrices. Then $$\Re(\det(R+iS))=\det(R )-\det(S).$$ $Proof.$ Indeed, if $R_1,R_2$ are the column vectors of $R$, and $S_1,S_2$ are the column vectors of $S$, then using the bi-linearity of the determinant we have \eqalign{ ... 9 This looks okay. Initially, we have that Av=\lambda v for eigenvalues \lambda of A. Since \lambda=0, we have that Av=0. Now assume that A^{-1} exists. Now by multiplying on the left by A^{-1}, we get v=0. This is a contradiction, since v cannot be the zero vector. So, A^{-1} does not exist. 7 The determinant is (up to sign and a factor of 6) the volume of a tetrahedron the origin and the three column vectors as vertices. Moving one vertex further from the plane determined by the other three increases this volume. By convexity, among the farthest points is at least one vertex. Thus the result for [-1,1] is the same as for \{-1,1\}. 6 In short, row reduce. In more detail: If two rows have their leftmost 1s in the same column, subtract the row with a shorter sequence of 1s from the row with the longer sequence. (If they're the same length, the rows are equal, so the determinant is zero and we're done.) This operation preserves the determinant and the condition that 1s only appear ... 6 If \lambda_1,\dotsc,\lambda_n are the (not necessarily distinct) eigenvalues of an n\times n matrix A, then \det(A)=\lambda_1\dotsb\lambda_n\tag{1} $$A nice proof of this fact can be found here. Now, A is invertible if and only if \det(A)\neq0. Hence (1) implies A is invertible if and only if 0 is not an eigenvalue of A. 4 Your proof works but I think could be cleaned up a bit. For example the sentence "We know A is an invertible and in order for A v=0; v=0" could be stated "For an invertible matrix A, Av=0 \Rightarrow v=0". You may need to justify that claim, perhaps by referring to an earlier result or just by pointing out that the map v \rightarrow Av is injective and ... 4 You have |\det[x_1 \cdots x_n ] | \le \|x_1\|\cdots \|x_n\| (Hadamard inequality). This shows the determinant is upper bounded by 16. Now choose A = \begin{bmatrix} 1 & 1 & 1 & 1 \\ 1 & -1 & 1 & -1 \\ 1 & 1 & -1 & -1 \\ 1 & -1 & -1 & 1 \end{bmatrix}, and compute \det A = 16. 4 Let $$A=\begin{bmatrix}I_{n-2} & \mathbf{0} & \mathbf{0} \\ \mathbf{0} & 0 & -1 \\ \mathbf{0} & 1 & 0\end{bmatrix},$$ then $$A^2=\begin{bmatrix} I_{n-2} & \mathbf{0} & \mathbf{0} \\\mathbf{0}&-1 & 0 \\ \mathbf{0}& 0 & -1\end{bmatrix}.$$ Now make B a matrix ... 4 Every real 3 \times 3 matrix, being of odd size, has a real eigenvalue since the characteristic polynomial is of odd degree. If A^2 + I = 0, and A\mathbf v = \lambda \mathbf v with \mathbf v \ne 0, then 0 = (A^2 + I) \mathbf v = (\lambda^2 + 1) \mathbf v \Rightarrow \lambda^2 + 1 = 0. Applying this notion to a real eigenvalue of A leads to an ... 3 Had the OP revealed the origin of the question, I might be able to devise a nicer solution. For the time being, I can only solve the problem by a brute-force approach. As M and N commute, there are only three possibilities: At least one of the two matrices is nonsingular. When M is nonsingular, the inequality is equivalent to$$4xz \det(xI + yNM^{-1} ...

3

Some rules about the determinant: if $A$ and $B$ are two square matrices with the same dimension $d$, $$\det(AB)=\det(A)\det(B).$$ In particular, since $\det(I)=1$, the determinant of the inverse of a matrix $A$ is the inverse of the determinant of $A$. For the multiplication by a scalar, $$\det(\lambda A)=\lambda^d\det(A)$$ since the determinant is a ...

3

Note that for $\lambda=a-b$ we have $$A-\lambda I = \begin{bmatrix} b & \cdots & b \\ \vdots & \ddots & \vdots \\ b & \cdots & b \end{bmatrix}\tag{1}$$ The matrix in $(1)$ has rank $1$ so its nullspace has dimension $n-1$. Hence $\lambda_1=a-b$ is an eigenvalue of $A$ whose geometric multiplicity is $n-1$. Now, note that ...

3

For any square matrix with one value on the diagonal and another value everywhere else, a consistent pattern of (orthogonal but not orthonormal) eigenvectors for the $n$ by $n$ case can be read from the columns of $$\left( \begin{array}{rrrrrrrrrr} 1 & -1 & -1 & -1 & -1 & -1 & -1 & -1 & -1 & -1 ... 3 Hint: use multilinearity of the determinant with respect to lines (or columns) and antisymetry. details:$$\det B =\det \begin{bmatrix} 2a & 2b & 2c \\ 4 & 4 & 4\\ 2 & 0 & 1\end{bmatrix} = 2\times 4\times \frac 12 \det \begin{bmatrix} a & b & c \\ 1 & 1 & 1\\ 4 & 0 & 2\end{bmatrix} = -4\det A = -12 \\ \det C ...

3

When $n$ is odd use the $n \times n$ matrix for the cyclic permutation $(a_1,a_2,...,a_n) \rightarrow (a_n,a_1,...,a_{n-1})$.. Then $A^n=I$ and since $n$ is odd, $det(A)=1$ and the entries of $A$ are all $1$ or $0$. The eigenvalues are the distinct $n$'th roots of unity (it has characteristic polynomial $x^n-1$). If you allow determinant $-1$ this will ...

3

You do not need the diagonal entries of $D$ to be positive. Since $\det(I+AB)=\det(I+BA)$ we have $\det(I+xy^T)=1+y^Tx$ (which is not hard to prove directly). If $D$ is invertible, then $$\det(D+ss^T) = \det(D(I+D^{-1}ss^T)) = \det(D) \det(I+D^{-1}ss^T) = (1+s^TD^{-1}s) \det(D).$$ This gives $$\det(D+ss^T) = \det(D) \left(1+\sum_r ... 3 Call your matrix H_n. For n\ge3, let P be the bidiagonal matrix whose main diagonal entries are 1 and whose superdiagonal entries are -1. Then \det(P)=1 and hence \det(H_n)=\det(PH_n). Since$$ PH_n=\pmatrix{ 2&-3\\ &3&-4\\ &&\ddots&\ddots\\ &&&\ddots&-(n-1)\\ &&&&n-1&-n\\ ...

3

Note that your matrix can be written as $$\text{diag}(a_1,a_2,\ldots,a_n) + \begin{bmatrix} 1\\1\\1\\ \vdots\\1\end{bmatrix} \begin{bmatrix} b_1 & b_2 & \cdots & b_n\end{bmatrix}$$ This is a rank $1$ update to a diagonal matrix, whose determinant can be computed using the Sylvester determinant theorem: $$\det(I+UV^T) = \det(I+V^TU)$$ I will leave ...

3

Your work is fine, but here's a little more to think about while moving forward in this subject. Multiplication by a square $n*n$ matrix produces a linear transformation (eg, translate, rotate, enlarge/shrink) in $n$ dimensions. If an eigenvalue is $0$, then that means that along the corresponding eigenvector, the transformation scales the image down to ...

2

Not sure how picky your Prof is about proofs, but I think your initial statement is backwards. The question is asking whether A is invertible given that it has an eigenvalue of 0. So you are trying to prove that, "If a square matrix A has an eigenvalue of 0, then A is NOT invertible." Thus, for the sake of contradiction you want to assume that A is ...

2

For your first question: $$I_n=\left(AA^{-1}\right)^T=\left(A^{-1}\right)^TA^T$$ so $A^T$ is invertible and its inverse is $\left(A^{-1}\right)^T$. For your last question: let $$f\colon \Bbb R^{11}\rightarrow \Bbb R^5,\quad x\mapsto A x$$ the linear transformation represented by the matrix $A$ in the standard basis, then by the rank-nullity theorem the ...

2

For your first question, I assume that you have discussed the property that $\left(AB\right)^{t} = B^t A^t$ for all square matrices $A$, $B$. If you know that $A$ is invertible, then you know that there exists a matrix $A^{-1}$ such that $AA^{-1} = I$. Now consider applying the property discussed. For your second question, it is tough to know exactly what ...

2

Your way is basically along the lines of how I would do it: Let $A$ be an $n \times n$ matrix, and assume for the sake of contradiction that $A$ is invertible. By your work, since $0$ is an eigenvalue of $A$, then we know there exists a nonzero vector $\mathbf{v}$ such that $A\mathbf{v} = 0$. $\Rightarrow dim(N(A)) > 0$, where $N(A)$ represents the ...

2

Here are two properties of determinants which I assume you know: (1) You can add/subtract two rows and determinant doesn't change. (2) If you multiply $k^{th}$ row by a constant, the determinant will be multiplied by the same constant. Suppose that the matrix $A$ has a row, say $k^{th}$ row, equal to sum of two other rows. Use law (1) and subtract those ...

2

$f : \sigma \in S_n \rightarrow f(\sigma) =\sigma^{-1} \in S_n$ is a bijection (you can trivially check it is its own inverse), and this is a general fact that for a finite sum $\sum_{i \in A}u_i = \sum_{j \in B}u_{f(j)}$ if $f$ is a bijection from $B$ to $A$ (actually it also works for some infinite sums, such as sums with only positive terms). As you ...

2

Do you know this definition? ($ϵ(σ):=$ the sign of the permutation) $$\det A = \sum_\sigma {\epsilon(\sigma)} \prod_i A_{i,\sigma (i)}$$ If you do: $$\det A^T = \sum_\sigma {\epsilon(\sigma)} \prod_i A_{\sigma (i),i} \\\forall\sigma\ \ \prod_i A_{\sigma(i),i} = \prod_j A_{j,\sigma^{-1}(j)}$$ As $\sigma \sigma^{-1} = {\rm id}, 1 = \epsilon( {\rm id})= ... 1 The problem with your approach is that generally in order to prove that$\det(AB) = \det(A) \det(B)$, one uses the fact that swapping two rows of a matrix multiplies the determinant by$-1$(see, for example, the second proof in http://www.proofwiki.org/wiki/Determinant_of_Matrix_Product). So you are kind of stuck proving the desired result directly. ... 1 I've found that the geometric sum of matrices is interesting, so look at$\sum_ { n = 0} ^ \infty A^n=(I-A)^{-1}$The series only converges if the largest eigenvalue of$ A $has absolute value less than one. An interesting conjecture would be that is relatively easy to calculate the determinant of$(I-A)^{-1}$but it would be harder to calculate the ... 1 Assume that line$k$is the sum of lines$i$and$j$and consider the vector$x$whose$i$and$j$entries are$-1$, entry$k$is$+2$and every other entry is$0$. Then$A^Tx=0$(can you show this?) and$x\ne0$hence$A^T$is not invertible, which implies that$\det(A^T)=0$. Since$\det(A)=\det(A^T)$(can you show this?), this shows that$\det(A)=0$. 1 Hints: Remember that eigenvectors of linear mappings are by definition non-zero. Take one eigenvector$\;v_i\;$from each different eigenvalue$\;\lambda_i\;$and prove this$\;n\;$eigenvectors are linearly independent and thus they are a basis for$\;V\;$. Calculate the matrix representation of$\;T\;$wrt the basis$\;\{v_1,...,v_n\}\;\$ . Further hint: ...

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