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Solving Ax=0A\mathbf{x}=\mathbf{0} : pivot variables, special solutions

讲座摘要（lecture summary）​

We have a definition for the column space and the nullspace of a matrix, but how do we compute these subspaces?

Computing the nullspace​

The nullspace of a matrix AA is made up of the vectors x\mathbf{x} for which Ax=0A\mathbf{x}=\mathbf{0}.

Suppose:

A=[1222246836810].A=\left[{\begin{array}{r r r r}{1}&{2}&{2}&{2}\\ {2}&{4}&{6}&{8}\\ {3}&{6}&{8}&{10}\end{array}}\right].

(Note that the columns of this matrix AA are not independent.) Our algorithm for computing the nullspace of this matrix uses the method of elimination, despite the fact that AA is not invertible. We don’t need to use an augmented matrix because the right side (the vector b) is 0 in this computation.

The row operations used in the method of elimination don’t change the solution to Ax=bA\mathbf{x}=\mathbf{b} so they don’t change the nullspace. (They do affect the column space.)

The first step of elimination gives us:

A=[1222246836810]⟶[122200240024].A={\left[\begin{array}{l l l l} {1}&{2}&{2}&{2}\\ {2}&{4}&{6}&{8}\\ {3}&{6}&{8}&{10}\end{array}\right]} \quad\longrightarrow\quad{\left[\begin{array}{l l l l} {1}&{2}&{2}&{2}\\ {0}&{0}&{2}&{4}\\ {0}&{0}&{2}&{4}\end{array}\right]}.

We don’t find a pivot in the second column, so our next pivot is the 22 in the third column of the second row:

[122200240024]⟶[122200240000]=U\left[{\begin{array}{r r r r}{1}&{2}&{2}&{2}\\ {0}&{0}&{2}&{4}\\ {0}&{0}&{2}&{4}\end{array}}\right] \quad\longrightarrow\quad\left[{ \begin{array}{r r r r} {1}&{2}&{2}&{2}\\ {0}&{0}&{2}&{4}\\ {0}&{0}&{0}&{0}\end{array}}\right]=U

The matrix UU is in echelon (staircase) form. The third row is zero because row 33 was a linear combination of rows 11 and 22; it was eliminated.

The rank of a matrix AA equals the number of pivots it has. In this example, the rank of AA (and of UU ) is 22.

Special solutions​

Once we’ve found UU we can use back-substitution to find the solutions x\mathbf{x} to the equation Ux=0U\mathbf{x}=\mathbf{0} . In our example, columns 11 and 33 are pivot columns containing pivots, and columns 22 and 44 are free columns. We can assign any value to x2x_{2} and x4;x_{4}; we call these free variables. Suppose x2=1x_{2}=1 and x4=0x_{4}=0 . Then:

2x3+4x4=0⟹x3=02x_{3}+4x_{4}=0\quad\Longrightarrow\quad x_{3}=0

and:

x1+2x2+2x3+2x4=0⟹x1=−2.x_{1}+2x_{2}+2x_{3}+2x_{4}=0\quad\Longrightarrow\quad x_{1}=-2.

So one solution is x=[−2100]\mathbf{x}=\left[\begin{array}{r}{-2}\\ {1}\\ {0}\\ {0}\end{array}\right] (because the second column is just twice the first column). Any multiple of this vector is in the nullspace.

Letting a different free variable equal 1 and setting the other free variables equal to zero gives us other vectors in the nullspace. For example:

x=[20−21]\mathbf{x}=\left[\begin{array}{r}{2}\\ {0}\\ {-2}\\ {1}\end{array}\right]

has x4=1x_{4}=1 and x2=0x_{2}=0 . The nullspace of AA is the collection of all linear combinations of these “special solution” vectors.

The rank rr of AA equals the number of pivot columns, so the number of free columns is n−rn-r : the number of columns (variables) minus the number of pivot columns. This equals the number of special solution vectors and the dimension of the nullspace.

Reduced row echelon form​

By continuing to use the method of elimination we can convert UU to a matrix RR in reduced row echelon form (rref form), with pivots equal to 1 and zeros above and below the pivots.

U=[122200240000]→[120−200240000]→[120−200120000]=R.U={\left[ \begin{array}{l l l l} {1}&{2}&{2}&{2}\\ {0}&{0}&{2}&{4}\\ {0}&{0}&{0}&{0}\end{array}\right]}\to{\left[ \begin{array}{l l l l} {1}&{2}&{0}&{-2}\\ {0}&{0}&{2}&{4}\\ {0}&{0}&{0}&{0}\end{array}\right]}\to{\left[ \begin{array}{l l l l}{1}&{2}&{0}&{-2}\\ {0}&{0}&{1}&{2}\\ {0}&{0}&{0}&{0}\end{array}\right]}=R.

By exchanging some columns, RR can be rewritten with a copy of the identity matrix in the upper left corner, possibly followed by some free columns on the right. If some rows of AA are linearly dependent, the lower rows of the matrix RR will be filled with zeros:

R=[IF00].R=\left[\begin{array}{l l}{I}&{F}\\ {0}&{0}\end{array}\right].

(Here II is an rr by rr square matrix.)

If NN is the nullspace matrix N=[−F I]N=\left[\begin{array}{l}{-F}\\ {~I}\end{array}\right] then RN=0R N=0 . (Here II is an n−rn-r by n−rn-r square matrix and 0 is an mm by n−rn-r matrix.) The columns of NN are the special solutions.

Problems and Solutions（习题及答案）​

Problem 7.1:

a) Find the row reduced form of:

A=[157904172−211−3]A={\left[\begin{array}{l l l l}{1}&{5}&{7}&{9}\\ {0}&{4}&{1}&{7}\\ {2}&{-2}&{11}&{-3}\end{array}\right]}

b) What is the rank of this matrix?

c) Find any special solutions to the equation Ax=0A\mathbf{x}=\mathbf{0} .

Solution

a) To transform AA into its reduced row form, we perform a series of row operations. Different operations are possible (same answer!). First, we multiply the first row by 2 and subtract it from the third row:

[157904172−211−3]⟶[157904170−12−3−21].\left[{\begin{array}{r r r r}{1}&{5}&{7}&{9}\\ {0}&{4}&{1}&{7}\\ {2}&{-2}&{11}&{-3}\end{array}}\right]\longrightarrow\left[{\begin{array}{r r r r}{1}&{5}&{7}&{9}\\ {0}&{4}&{1}&{7}\\ {0}&{-12}&{-3}&{-21}\end{array}}\right].

We then multiply the second row by 14\frac14 to make the second pivot 1:

[157904170−12−3−21]⟶[1579011/47/40−12−3−21].\left[{\begin{array}{r r r r}{1}&{5}&{7}&{9}\\ {0}&{4}&{1}&{7}\\ {0}&{-12}&{-3}&{-21}\end{array}}\right]\longrightarrow\left[{\begin{array}{r r r r}{1}&{5}&{7}&{9}\\ {0}&{1}&{1/4}&{7/4}\\ {0}&{-12}&{-3}&{-21}\end{array}}\right].

Multiply the second row by 12 and add it to the third row:

[1579011/47/40−12−3−21]⟶[1579011/47/40000] .{\left[\begin{array}{l l r r}{1}&{5}&{7}&{9}\\ {0}&{1}&{1/4}&{7/4}\\ {0}&{-12}&{-3}&{-21}\end{array}\right]}\longrightarrow{\left[\begin{array}{l l r r}{1}&{5}&{7}&{9}\\ {0}&{1}&{1/4}&{7/4}\\ {0}&{0}&{0}&{0}\end{array}\right]}\,.

Finally, multiply the second row by 5 and subtract it from the first row:

[1579011/47/40000]⟶[10  23/41/401  1/47/400   00]{\left[\begin{array}{l l l l}{1}&{5}&{7}&{9}\\ {0}&{1}&{1/4}&{7/4}\\ {0}&{0}&{0}&{0}\end{array}\right]}\longrightarrow{\left[\begin{array}{l l l l}{1}&{0}&{\ \ 23/4}&{1/4}\\ {0}&{1}&{\ \ 1/4}&{7/4}\\ {0}&{0}&{\ \ \ 0}&{0}\end{array}\right]}

b) The matrix is of rank 2 because it has 2 pivots.

c) The special solutions to Ax=0A\mathbf{x}=\mathbf{0} are:

[−23/4−1/410] and [−1/4−7/401]\left[{\begin{array}{r}{-23/4}\\ {-1/4}\\ {1}\\ {0}\end{array}}\right]{\mathrm{~and~}}\left[{\begin{array}{r}{-1/4}\\ {-7/4}\\ {0}\\ {1}\end{array}}\right]

Problem 7.2: (3.3 #17.b Introduction to Linear Algebra: Strang) Find A1A_{1} and A2A_{2} so that rank (A1B)=1(A_{1}B)=1 and rank (A2B)=0(A_{2}B)=0 for B=[1111]B={\left[\begin{array}{l l}{1}&{1}\\ {1}&{1}\end{array}\right]}

Solution

Solution: Take A1=I2A_{1}=I_{2} and A2=02A_{2}=0_{2} . A less trivial example is A2=[1−11−1]A_{2}={\left[\begin{array}{l l}{1}&{-1}\\ {1}&{-1}\end{array}\right]} .