# kernel.pdf - y 3 11 v 11 = 8 v 517 v 11 = 5 v 47 1 and...

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How to find a basis of the kernel/null space?Stat 155, Game Theory (Fall 2017)Soumendu Sundar MukherjeeSeptember 27, 2017Note:If you have comments or questions about the solutions, feel free to emailme at[email protected].Many of you asked in today’s section how the two basis vectors for the nullspace of the matrixB=8341-14716-10385-1was derived in the handouts.Well,Bz= 0 is an underdetermined system inthat we have more variables (five) than equations (three).So what you cando is to treat two of the variables as paramaters, and solve for the other threevariables in terms of these parameters. In the handouts,y4, andvwere treatedas parameters. Once you row-reduce, the system becomes13/81/21/8-1/801-2/111-1/1100111/47-4/47y1y2y3y4v=000.Note that since we have two free parametersy4andv, the dimension of thisspace is going to be 2, i.e. there will be two basis vectors. We are free to choosethese parameters. A convenient choice is to set each of them to zero separately.If we sety4= 0, and solve fory1, y2, y3, we get from the third row thaty3= 4v/47,from the second row that