• Linear Algebra

    How to compute basis for the range, nullspace etc. of a matrix? 6 Approaches

    The four fundamental spaces of a matrix $A$, namely the range and the nullspace of itself $A$ or its transpose $A^T$, are the heart of linear algebra. We often find ourselves in need of computing a basis for the range or the nullspace of a matrix, for theoretical or applicational purposes. There are many ways of computing a basis for the range or nullspace for $A$ or $A^T$. Some are better for application, either due to their robustness against floating point…

  • Linear Algebra

    How to prove SVD ? A recipe approach

    To prove the existence of SVD is no trivial task, but it turns out that it’s not too difficult either. Looks like one needs a few ingredients (hence the title), but once we know them and understand the overall idea, the proof is not too difficult. Below we list the basic ingredients needed to prove the existence of SVD. The URV decomposition $\mathbf{A} = \mathbf{URV} = \mathbf{U}\begin{pmatrix}\mathbf C & \mathbf 0 \\ \mathbf 0 & \mathbf 0\end{pmatrix}\mathbf{V}$ $||\mathbf{A}||_2 = ||\mathbf{URV}||_2 =…

  • Linear Algebra

    How can we compute basis for Nullspace?

    There are at least two answers to this question; one of these is more educative and the other one is at least as educative (in a different and profound way) as well as practical. Method 1 The first method is a more introductory level method. It is helpful to know it and good to read it as a refresher even if one is more advanced student of the topic. A linear system $\mathbf{Ax=b}$ is homogeneous when $\mathbf b=\mathbf 0$ and nonhomogeneous…

  • Linear Algebra

    Since we already have SVD, do we need URV factorization?

    The SVD factorization is a special case of the URV factorization that has many great properties. The latter decomposes a matrix $A_{m\times n}$ as $$A=URV^T,$$ where $U=(U_1|U_2)$ is an orthonormal matrix such that $U_1$ is a basis for $R(A)$, $U_2$ is a basis for $N(A^T)$; and $V=(V_1|V_2)$ is another orthonormal matrix such that $V_1$ is a basis for $R(A^T)$ and $V_2$ is a basis for $N(A)$. Suppose that the rank of $A$ is $r$, in which case $U_1$ has $r$ columns…