Linear Algebra
http://www.maa.org/taxonomy/term/42292/0
enImage Reconstruction in Linear Algebra
http://www.maa.org/programs/faculty-and-departments/classroom-capsules-and-notes/image-reconstruction-in-linear-algebra
<div class="field field-name-body field-type-text-with-summary field-label-hidden"><div class="field-items"><div class="field-item even"><p><em>The main object is to solve the inverse problem of recovering the original scene, represented by a vector or a matrix, from its photograph, represented by a product of a matrix and the original vector or matrix. The solution of the resulting matrix equation gives rise to the reconstruction of the original scene</em>.</p>
</div></div></div>Fri, 12 Jul 2013 15:03:25 +0000newton_admin95016 at http://www.maa.orgA Matrix Proof of Newton's Identities
http://www.maa.org/programs/faculty-and-departments/classroom-capsules-and-notes/a-matrix-proof-of-newtons-identities
<div class="field field-name-body field-type-text-with-summary field-label-hidden"><div class="field-items"><div class="field-item even"><p><em>Newton’s identities relate the coefficients of a polynomial to sums of powers of its roots. The author uses the Cayley-Hamilton theorem and properties of the trace of a matrix to derive Newton’s identities.</em></p>
</div></div></div>Fri, 12 Jul 2013 15:03:25 +0000newton_admin95528 at http://www.maa.orgDoes the Generalized Inverse of \(A\) Commute with \(A\)?
http://www.maa.org/programs/faculty-and-departments/classroom-capsules-and-notes/does-the-generalized-inverse-of-a-commute-with-a
<div class="field field-name-body field-type-text-with-summary field-label-hidden"><div class="field-items"><div class="field-item even"><p><em>Not all square matrices commute with their generalized inverse (Moore-Penrose inverse). The author gathers equivalent conditions for the generalized inverse of a matrix to commute with the matrix itself. Then he shows that, in this case, the generalized inverse may be represented as a polynomial in the given matrix.</em></p>
</div></div></div>Fri, 12 Jul 2013 15:03:25 +0000newton_admin95504 at http://www.maa.orgProof without Words: A 2 x 2 Determinant Is the Area of a Parallelogram
http://www.maa.org/programs/faculty-and-departments/classroom-capsules-and-notes/proof-without-words-a-2-x-2-determinant-is-the-area-of-a-parallelogram
<div class="field field-name-body field-type-text-with-summary field-label-hidden"><div class="field-items"><div class="field-item even"><p><em>The author presents a visual proof that the determinant of a 2 by 2 matrix equals the area of the corresponding parallelogram.</em></p>
</div></div></div>Fri, 12 Jul 2013 15:03:25 +0000newton_admin95382 at http://www.maa.orgComputing the Fundamental Matrix for a Reducible Markov Chain
http://www.maa.org/programs/faculty-and-departments/classroom-capsules-and-notes/computing-the-fundamental-matrix-for-a-reducible-markov-chain
<div class="field field-name-body field-type-text-with-summary field-label-hidden"><div class="field-items"><div class="field-item even"><p><em>A Markov chain with 9 states is used to illustrate a technique for finding the fundamental matrix.</em></p>
</div></div></div>Fri, 12 Jul 2013 15:03:25 +0000newton_admin95376 at http://www.maa.orgA Note on the Equality of the Column and Row Rank of a Matrix
http://www.maa.org/programs/faculty-and-departments/classroom-capsules-and-notes/a-note-on-the-equality-of-the-column-and-row-rank-of-a-matrix
<div class="field field-name-body field-type-text-with-summary field-label-hidden"><div class="field-items"><div class="field-item even"><p><em>An elementary argument, different from the usual one, is given for the familiar equality of row and column rank. The author takes “full advantage of the following two elementary observations: (1) For any vector \(x\) in \(\mathcal{R}^n\) and matrix \(A\), \(Ax\) is a linear combination of the columns of \(A\), and (2) vectors in the null space of \(A\) are orthogonal to vectors in the row space of \(A\), relative to the usual Euclidean product.”</em></p>
</div></div></div>Fri, 12 Jul 2013 15:03:25 +0000newton_admin95365 at http://www.maa.orgThe Axis of a Rotation: Analysis, Algebra, Geometry
http://www.maa.org/programs/faculty-and-departments/classroom-capsules-and-notes/the-axis-of-a-rotation-analysis-algebra-geometry
<div class="field field-name-body field-type-text-with-summary field-label-hidden"><div class="field-items"><div class="field-item even"><p><em>Given any 3 by 3 rotation matrix \(A\) (i.e. orthogonal with determinant \(1\) and an arbitrary vector \(x\), the vector \( Ax +A^{T}x+[1−\)tr\((A)]x\) lies on the axis of rotation. The article provides three different approaches, requiring various levels of background knowledge, to prove and/or explain the given result.</em></p>
</div></div></div>Fri, 12 Jul 2013 15:03:25 +0000newton_admin95339 at http://www.maa.orgTwo by Two Matrices with Both Eigenvalues in \(Z/pZ\)
http://www.maa.org/programs/faculty-and-departments/classroom-capsules-and-notes/two-by-two-matrices-with-both-eigenvalues-in-zpz
<div class="field field-name-body field-type-text-with-summary field-label-hidden"><div class="field-items"><div class="field-item even"><p><em>This article provides a non-group theory approach to finding the number of two by two matrices over \( Z/pZ\) that have both eigenvalues in the same field. The strategy is to use the quadratic formula to find the roots of the characteristic polynomial of a matrix and then count the number of matrices for which these roots are in \(Z/pZ \).</em></p>
</div></div></div>Fri, 12 Jul 2013 15:03:25 +0000newton_admin95269 at http://www.maa.orgA Geometric Approach to Linear Functions
http://www.maa.org/programs/faculty-and-departments/classroom-capsules-and-notes/a-geometric-approach-to-linear-functions
<div class="field field-name-body field-type-text-with-summary field-label-hidden"><div class="field-items"><div class="field-item even"><p><em>There are three somewhat distinct topics in this article: the condition for linear functions to commute, a linear function as a transformation of the number line, and linear difference equations. A linear function \(y=f(x)=ax+b\) can be characterized in terms of slope and the “center of reflection,” both of which reflect the geometric property of the function. </em></p>
</div></div></div>Fri, 12 Jul 2013 15:03:25 +0000newton_admin95221 at http://www.maa.orgThe Arithmetic of Algebraic Numbers: An Elementary Approach
http://www.maa.org/programs/faculty-and-departments/classroom-capsules-and-notes/the-arithmetic-of-algebraic-numbers-an-elementary-approach
<div class="field field-name-body field-type-text-with-summary field-label-hidden"><div class="field-items"><div class="field-item even"><p><em>If \(r\) and \(s\) are algebraic numbers, then \(r + s\), \(rs\), and \(r/s\) are also algebraic. The proof provided in this capsule uses the ideas of characteristic polynomials, eigenvalues, and eigenvectors.</em></p>
</div></div></div>Fri, 12 Jul 2013 15:03:25 +0000newton_admin95166 at http://www.maa.orgThe Existence of Multiplicative Inverses
http://www.maa.org/programs/faculty-and-departments/classroom-capsules-and-notes/the-existence-of-multiplicative-inverses
<div class="field field-name-body field-type-text-with-summary field-label-hidden"><div class="field-items"><div class="field-item even"><p><em>Using only basic ideas from linear algebra and number theory, the authors show that if \(c\) is square-free, the ring \(Q [\sqrt[n]{c}] \) is a field. An arbitrary nonzero element of the ring is associated with a system of equations, and divisibility arguments are used to show that a matrix of coefficients from the system must have a nonzero determinant, eventually leading to the result that the original element of the ring has an inverse. </em></p>
</div></div></div>Fri, 12 Jul 2013 15:03:25 +0000newton_admin95149 at http://www.maa.orgNotational Collisions
http://www.maa.org/programs/faculty-and-departments/classroom-capsules-and-notes/notational-collisions
<div class="field field-name-body field-type-text-with-summary field-label-hidden"><div class="field-items"><div class="field-item even"><p>This capsule points out several potential confusions in commonly used linear algebra notation.</p>
</div></div></div>Fri, 12 Jul 2013 15:03:25 +0000newton_admin95133 at http://www.maa.orgUsing Quadratic Forms to Correct Orientation Errors in Tracking
http://www.maa.org/programs/faculty-and-departments/classroom-capsules-and-notes/using-quadratic-forms-to-correct-orientation-errors-in-tracking
<div class="field field-name-body field-type-text-with-summary field-label-hidden"><div class="field-items"><div class="field-item even"><p><em>If noise in data transmission produces a not quite orthogonal matrix that is known to be orthogonal, how does one find the "nearest" orthogonal matrix? This capsule recasts the problem as one of maximizing a quadratic form on the four-dimensional unit sphere, and sketches a solution.</em></p>
</div></div></div>Fri, 12 Jul 2013 15:03:25 +0000newton_admin95396 at http://www.maa.orgA Polynomial Taking Integer Values
http://www.maa.org/programs/faculty-and-departments/classroom-capsules-and-notes/a-polynomial-taking-integer-values
<div class="field field-name-body field-type-text-with-summary field-label-hidden"><div class="field-items"><div class="field-item even"><p><em>The article supplies a short, elementary proof that for integers \(a_1 < a_2 < \cdots < a_n \), the expression \( \prod_{n \geq i > j \geq 1} \frac{a_i - a_j}{i-j} \) is an integer. This previously known result is proved using the Vandermonde determinant. (Please note a typo in the first sentence of the paper where a fraction bar has been omitted.)</em></p>
</div></div></div>Fri, 12 Jul 2013 15:03:25 +0000newton_admin95399 at http://www.maa.orgDeterminants of the Tournaments
http://www.maa.org/programs/faculty-and-departments/classroom-capsules-and-notes/determinants-of-the-tournaments
<div class="field field-name-body field-type-text-with-summary field-label-hidden"><div class="field-items"><div class="field-item even"><p><em>Consider a tournament with \(n\) players where each player plays every other player once, and ties are not allowed. An \( n \times n\) tournament matrix \(A\) is constructed where diagonal entries are zero, \(A_{ij} = 1\) if \(i\) beats \(j\), and \(A_{ij}=-1\) if \(j\) beats \(i\). The authors demonstrate that the determinant of a tournament matrix is zero if and only if \(n\) is odd. Additionally, it is shown that the nullspace of a tournament matrix has dimension zero if \(n\) is even and dimension one if \(n\) is odd.</em></p>
</div></div></div>Fri, 12 Jul 2013 15:03:25 +0000newton_admin95402 at http://www.maa.org