# Tag Info

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No. Consider the $2 \times 2$ Jordan block $$\pmatrix{0 & 1 \\ 0 & 0},$$ or the matrix $$\pmatrix{0 & -1 \\ 1 & 0},$$ which represents an anticlockwise rotation by $\frac{\pi}{2}$.

3

The short answer: Because you multiply nonsingular matrices from the left. The long answer: Say you have a matrix $A$. Each Gaussian elimination step corresponds to some elementary matrix (which are nonsingular). Thus, there exists a nonsingular matrix $M$ (product of elementary matrices) such that $MA$ has reduced row echelon form. Now, let $x$ be in the ...

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Say we have an $n$x$n$ matrix, $A$, and are going to row reduce it. Every time we do a row operation, it is the same as multiplying on the left side by an invertible matrix corresponding to the operation. So at the end of the process, we can conclude something like $B = L_1L_2...L_kA$, where $B$ is the row reduced matrix, and the $L_i$ are the matrices ...

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The operations (elementary row operations) that occur in Gauss-Jordan elimination for the purposes of row reduction are mathematically equivalent to left multiplication by invertible matrices known as elementary matrices; these in fact generate the general linear group of $n\times n$ invertible real matrices. Because elementary matrices are invertible, it ...

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This is because, interpreting the rows of the matrix as a system of linear equations, the original matrix and its row-reduced form correspond to logically equivalent systems. Indeed we can go back to the original matrix (the original system of equations) by means of the inverse transformations on rows.

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No you can take the nilpotent matrix $$M=\left(\begin{array}{cccc} 0 & 1 & 0& 0\\ 0&0 & 1& 0\\ 0& 0&0&1\\ 0& 0& 0& 0 \end{array} \right)$$ M isn't symmetrical but $A^4=0_4$ symmetric

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Only square matrices have determinants. I'm not sure whether your matrix was square to begin with, but it's not square both before and after adding that extra row. If $Ax=0$ has a unique solution, $B$ is the matrix obtained from $A$ by adding another row, and that row is a linear combination of the rows of $A$ then yes, $Bx=0$ has a unique solution. In fact ...

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The first question only requires the definition of an eigenvalue. For the second one, since $g\in \mathbb{C}[X], g(X)=c\prod\limits_{i=1}^n(X-\lambda_i)$ so $g(A)=c\prod\limits_{i=1}^n(A-\lambda_i I)$. And since the product of invertible matrices is invertible, you can conclude using the first point.

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I don't have Hoffman & Kunze at hand, but the definition of row echelon form (which is embedded in your definition of reduced row echelon form) from wikipediae states: The leading coefficient (the first nonzero number from the left, also called the pivot) of a nonzero row is always strictly to the right of the leading coefficient of the row ...

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It's probably because it's not a geometrically meaningful operation; a linear transformation whose matrix in one basis is all ones, has another matrix in another basis. Whenever I've seen the notation $A+b$ in mathematics, it has meant $A+bI$ (where $A$ is a quadratic matrix and $I$ is the identity matrix of the same size). Some people write ...

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Yes, that relationship is true. $A^2$, $A$, and $I$ can all be simultaneously diagonalized by an orthogonal matrix $E$. If $\Lambda$ are the eigenvalues of $A$ then we know that $$A^2 = E\Lambda^2 E^{-1} \,\,\,\, A = E\Lambda E^{-1} \,\,\,\, I = E I E^{-1}$$ The question of whether the ordering you specified is true is by definition the question of whether ...

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If you take an initial point $p_i$, apply a transform $A$, then apply transform $B$, then apply transform $C$ to get a final point $p_f$, then we have $$p_f = C(B(A(p_i)))$$ $$p_f = CBAp_i$$ So the total transformation is $t = CBA$. This is one of the reasons why matrix multiplication is so nice--it lets us compose many linear transforms into a single ...

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Each 4×4 transformation matrix is of the form $$T_i = \begin{vmatrix} R_i & \vec{t}_i \\ 0 & 1 \end{vmatrix}$$ such that when applied to a 4×1 point $(\vec{p},1)$ the result is a rotation and a translation $(\vec{t}_i+R_i \vec{p},1)$ Now combine three transformations for $$\begin{vmatrix} R_1 & \vec{t}_1 \\ 0 & 1 \end{vmatrix} ... 1 In general you should investigate on iterative methods to solve Ax=b. If A is not positive definite, there are provable no cheap methods. Rate of convergence and effective complexity depends strongly on the structure of A and preconditioning is almost a must. There are fix point iteration based methods and Krylov space based methods (like CG). A good ... 1 Firstly, we have that \dim\ker(\mathbf A-4\mathbf I) = 1. That means the eigenvalue \lambda = 4 is not semi - simple *, thus we have to add a unit above the main diagonal on the Jordan block that corresponds to the eigenvalue \lambda = 4, i.e.$$\mathbf J = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 4 & 1 \\ 0 & 0 & 4 \end{bmatrix}. ...

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