Specifically, I will optimize a matrix transpose to show how to use shared memory to reorder strided global memory accesses into coalesced accesses. In this post I will show some of the performance gains achievable using shared memory. My previous CUDA Fortran post covered the mechanics of using shared memory, including static and dynamic allocation. Therefore, we have constructed a full-rank set of eigenvectors of A, meaning that it is diagonalizable.CUDA Fortran for Scientists and Engineers shows how high-performance application developers can leverage the power of GPUs using Fortran. This shows that if x is an eigenvector of $A^TA$ then it is also an eigenvector of $A^2$, which in turn means it is an eigenvector of $A$ since powers of matrices share the same eigenspace. Applying A to both sides of this equation, we have $$A^2x=A^TAx=\lambda x$$. Subtracting these two equations, we have $$((A^T-A)x,(A^T-A)x)=0,$$ which by the definition of inner product implies that $$(A^T-A)x=0 =>A^Tx=Ax.$$ Note that of course this is true whenever $A=A^T$ if A is symmetric, but is a more general condition, since x is an eigenvector, rather than an arbitrary vector. Then $$(Ax,Ax)=(A^TAx,x)=\lambda(x,x)$$ and similarly $$(A^Tx,A^Tx)=(AA^Tx,x)=\lambda(x,x)$$ where I have used the normality of A. To see that $A$ normal implies $A^TA$ is diagonalizable, let $\lambda$ be a eigenvalue of $A^TA$ corresponding to the eigenvector x. As an obvious special case, $A$ is normal if $A$ is Hermitian (symmeric in the real case). In this case, we have that $A$ is diagonalizable. Note this is a stronger condition than saying that $A^TA$ is symmetric, which is always true. Something that occurred to me while reading this answer for help with my homework is that there is a pretty common and important special case, if the linear operator A is normal, i.e. Least Squares methods (employing a matrix multiplied with its transpose) are also very useful withĪutomated Balancing of Chemical Equations How can we use this routine for inverting an arbitrary matrix $A$ ?Īssuming that the inverse $A^\right] Suppose that we have a dedicated matrix inversion routine at our disposal, namely for a matrix $A$ Another interesting application of the specialty of $A^TA$ is perhaps the following. Any employment for the Varignon parallelogram?ĮDIT.What is the difference between Finite Difference Methods,įinite Element Methods and Finite Volume Methods for solving PDEs?.Solving for streamlines from numerical velocity field.Scale vector in scaled pivoting (numerical methods).Special? Yes! The matrix $A^TA$ is abundant with the Least Squares Finite Element Method in Numerical Analysis: Note, that the resulted eigenvectors are not yet normalized. Now, the originally searched eigenvectors $v_i$ of $AA^T$ can easily be obtained by $v_i:=Au_i$. Pre-multiplying both sides of this equation with $A$ yields In case $A$ is not a square matrix and $AA^T$ is too large to efficiently compute the eigenvectors (like it frequently occurs in covariance matrix computation), then it's easier to compute the eigenvectors of $A^TA$ given by $A^TAu_i = \lambda_i u_i$. The eigenvector decomposition of $AA^T$ is given by $AA^Tv_i = \lambda_i v_i$. Moreover, we can infer the eigenvectors of $A^TA$ from $AA^T$ and vice versa. That the rank is identical implies that the number of non-zero eigenvectors is identical. Indeed, independent of the size of $A$, there is a useful relation in the eigenvectors of $AA^T$ to the eigenvectors of $A^TA$ based on the property that $rank(AA^T)=rank(A^TA)$. The rest of the eigenvectors are the null space of $A$ i.e. $AA^T$ is positive semi-definite, and in a case in which $A$ is a column matrix, it will be a rank 1 matrix and have only one non-zero eigenvalue which equal to $A^TA$ and its corresponding eigenvector is $A$.
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