# “Don’t invert that matrix” – why and how

The first time I read John Cook’s advice “Don’t invert that matrix,” I wasn’t sure how to follow it. I was familiar with manipulating matrices analytically (with pencil and paper) for statistical derivations, but not with implementation details in software. For reference, here are some simple examples in MATLAB and R, showing what to avoid and what to do instead.

[Edit: R code examples and results have been revised based on Nicholas Nagle’s comment below and advice from Ryan Tibshirani.]

If possible, John says, you should just ask your scientific computing software to directly solve the linear system $Ax = b$. This is often faster and more numerically accurate than computing the matrix inverse of A and then computing $x = A^{-1}b$.

We’ll chug through a computation example below, to illustrate the difference between these two methods. But first, let’s start with some context: a common statistical situation where you may think you need matrix inversion, even though you really don’t.

[One more edit: I’ve been guilty of inverting matrices directly, and it’s never caused a problem in my one-off data analyses. As Ben Klemens comments below, this may be overkill for most statisticians. But if you’re writing a package, which many people will use on datasets of varying sizes and structures, it may well be worth the extra effort to use solve or QR instead of inverting a matrix if you can help it.]

(Be aware that the $x$ above (from traditional linear systems notation), and the $X$ below (from traditional regression notation), play totally different roles! It’s a shame that these notations conflict.)

Statistical context: linear regression

First of all, I’m used to reading and writing mathematical/statistical formulas using inverted matrices. In statistical theory courses, for example, we derive the equations behind linear regression this way all the time. If your regression model is $y = X \beta + \epsilon$, with independent errors of mean 0 and variance $\sigma^2$, then textbooks usually write the ordinary least squares (OLS) solution as $\hat{\beta} = (X'X)^{-1}X'y$.
Computing $\hat{\beta}$ directly like this in R would be
beta = solve(t(X) %*% X) %*% (X %*% y)
while in MATLAB it would be
beta = inv(X' * X) * (X' * y)

This format is handy for deriving properties of OLS analytically. But, as John says, it’s often not the best way to compute $\hat{\beta}$. Instead, rewrite this equation as $(X'X) \beta = X'y$ and then get your software to solve for $\beta$.
In R, this would be
beta = solve(t(X) %*% X, t(X) %*% y)
or in MATLAB, it would be
beta = (X' * X) \ (X' * y)

(Occasionally, we do actually care about the values inside an inverted matrix. Analytically we can show that, in OLS, the variance-covariance matrix of the regression coefficients is $Var(\hat{\beta}) = \sigma^2 (X'X)^{-1}$. If you really need to report these variances and covariances, I suppose you really will have to invert the matrix. But even here, if you only need them temporarily as input to something else, you can probably compute that “something else” directly without matrix inversion.)

Numerical example of problems with matrix inversion

The MATLAB documentation for inv has a nice example comparing computation times and accuracies for the two approaches.

Reddit commenter five9a2 gives an even simpler example in Octave (also works in MATLAB).

Here, I’ll demonstrate five9a2’s example in R. We’ll use their same notation of solving the system $Ax=b$ (rather than the regression example’s notation). We’ll let A be a 7×7 Hilbert matrix. The Hilbert matrices, with elements $H(i,j) = 1/(i+j-1)$, are known to be poorly conditioned [1] and therefore to cause trouble with matrix inversion.

Here’s the R code and results, with errors and residuals defined as in the MATLAB example:

set.seed(13052015)
options(digits = 3)
library(Matrix)
library(matrixcalc)

# Set up a linear system Ax=b,
# and compare
# inverting A to compute x=inv(A)b
# vs
# solving Ax=b directly

# Generate the 7x7 Hilbert matrix
# (known to be poorly conditioned)
n = 7
(A = as.matrix(Hilbert(n)))
##       [,1]  [,2]  [,3]  [,4]   [,5]   [,6]   [,7]
## [1,] 1.000 0.500 0.333 0.250 0.2000 0.1667 0.1429
## [2,] 0.500 0.333 0.250 0.200 0.1667 0.1429 0.1250
## [3,] 0.333 0.250 0.200 0.167 0.1429 0.1250 0.1111
## [4,] 0.250 0.200 0.167 0.143 0.1250 0.1111 0.1000
## [5,] 0.200 0.167 0.143 0.125 0.1111 0.1000 0.0909
## [6,] 0.167 0.143 0.125 0.111 0.1000 0.0909 0.0833
## [7,] 0.143 0.125 0.111 0.100 0.0909 0.0833 0.0769
# Generate a random x vector from N(0,1)
x = rnorm(n)
# Find the corresponding b
b = A %*% x

# Now solve for x, both ways,
# and compare computation times
system.time({xhat_inverting = solve(A) %*% b})
##    user  system elapsed
##   0.002   0.002   0.032
system.time({xhat_solving = solve(A, b)})
##    user  system elapsed
##   0.001   0.000   0.001

# Compare errors: sum of squared (x - xhat)
(err_inverting = norm(x - xhat_inverting))
## [1] 2.44e-07
(err_solving = norm(x - xhat_solving))
## [1] 1.56e-08

# Compare residuals: sum of squared (b - bhat)
(res_inverting = norm(b - A %*% xhat_inverting))
## [1] 1.55e-08
(res_solving = norm(b - A %*% xhat_solving))
## [1] 2.22e-16

As you can see, even with a small Hilbert matrix:

• inverting takes more time than solving;
• the error in x when solving Ax=b directly is a little smaller than when inverting; and
• the residuals in the estimate of b when solving directly are many orders of magnitude smaller than when inverting.

Repeated reuse of QR or LU factorization in R

Finally: what if, as John suggests, you have to solve Ax=b for many different b’s? How do you encode this in R without inverting A?

I don’t know the best, canonical way to do this in R. However, here are two approaches worth trying: the QR decomposition and the LU decomposition. These are two ways to decompose the matrix A into factors with which it should be easier to solve $Ax=b$. (There are other decompositions too—many more than I want to go into here.)

QR decomposition is included in base R. You use the function qr once to create a decomposition, store the Q and R matrices with qr.Q and qr.R, then use a combination of backsolve and matrix multiplication to solve for x repeatedly using new b’s. (Q is chosen to be orthogonal, so we know its inverse is just its transpose. This avoids the usual problems with matrix inversion.)

For the LU decomposition, we can use the matrixcalc package. (Thanks to sample code on the Cartesian Faith blog.)
Imagine rewriting our problem as $Ax = LUx = b$, then defining $y=Ux$, so that we can solve it in two stages: $Ly = b, Ux = y$. We can collapse this in R into a single line, in the form
x = backsolve(U, forwardsolve(L, b))
once we have decomposed A into L and U.

# What if we need to solve Ax=b for many different b's?
# Compute the A=QR or A=LU decomposition once,
# then reuse it with different b's.

# Just once, on the same b as above:

# QR, Ryan's suggestion:
system.time({
qr_decomp = qr(A, tol = 1e-10)
qr_R = qr.R(qr_decomp)
qr_Qt = t(qr.Q(qr_decomp))
xhat_qr = backsolve(qr_R, qr_Qt %*% b)
})
##    user  system elapsed
##   0.013   0.000   0.014
(err_qr = norm(x - xhat_qr))
## [1] 5.78e-08
(res_qr = norm(b - A %*% xhat_qr))
## [1] 9.44e-16

# LU, Nicholas' suggestion:
system.time({
lu_decomp = lu.decomposition(A)
xhat_lu = backsolve(lu_decomp$U, forwardsolve(lu_decomp$L, b))
})
##    user  system elapsed
##   0.003   0.000   0.016
(err_lu = norm(x - xhat_lu))
## [1] 3.01e-08
(res_lu = norm(b - A %*% xhat_lu))
## [1] 3.33e-16

Both the QR and LU decompositions’ errors and residuals are comparable to the use of solve earlier in this post, and much smaller than inverting A directly.

However, what about the timing? Let’s create many new b vectors and time each approach: QR, LU, direct solving from scratch (without factorizing A), and inverting A:

# Reusing decompositions with many new b's:
m = 1000
xs = replicate(m, rnorm(n))
bs = apply(xs, 2, function(xi) A %*% xi)

# QR, Ryan's suggestion:
system.time({
qr_decomp = qr(A, tol = 1e-10)
qr_R = qr.R(qr_decomp)
qr_Qt = t(qr.Q(qr_decomp))
xhats_qr = apply(bs, 2,
function(bi) backsolve(qr_R,
qr_Qt %*% bi))
})
##    user  system elapsed
##   0.036   0.000   0.036

# LU, Nicholas' suggestion:
system.time({
lu_decomp = lu.decomposition(A)
bs = apply(bs, 2,
function(bi) backsolve(lu_decomp$U, forwardsolve(lu_decomp$L, bi)))
})
##    user  system elapsed
##   0.094   0.001   0.095

# Compare to repeated use of solve()
system.time({
xhats_solving = apply(bs, 2,
function(bi) solve(A, bi))
})
##    user  system elapsed
##   0.082   0.000   0.082

# Compare to inverting A directly
system.time({
Ainv = solve(A)
xhats_inverting = apply(bs, 2,
function(bi) Ainv %*% bi)
})
##    user  system elapsed
##    0.01    0.00    0.01

It’s not surprising that inverting A is the fastest… but as we saw above, it’s also the least accurate, by far. Luckily, a well-done QR is almost as fast, and far more accurate.

However, I’m surprised that QR and LU are both much LU is a little slower than direct use of solve(A, b). They are LU is supposed to save you work that solve(A, b) has to redo for every new b.

Please comment if you know why this happened! Did I make a mistake? Or does LU and QR give you speedups over solve(A, b) only for much larger A matrices?

[Edit: Nicholas’ comment below and an email from Ryan Tibshirani show how to speed up the QR and LU approaches considerably. Even so, both are LU is still slower than using solve directly. See [2] for the earlier (slower) QR and LU code and results.]

Finally, see this Revolutions post on R and Linear Algebra for more on matrix manipulation in R. They mention dealing with giant and/or sparse matrices, which is also the last situation described in John Cook’s blog post.

[1] (I usually hear of a “poorly conditioned” matrix, meaning one with a high “condition number,” being defined in terms of the ratio of largest to smallest eigenvalues. However, this nice supplement on Condition Numbers from Lay’s Linear Algebra has a more general definition on p.4: if A is invertible, the condition number is the norm of A times the norm of A’s inverse. This is the same as the ratio of largest to smallest eigenvalues if you’re using the spectral norm… but the general definition is more interpretable for beginners who haven’t studied eigenvalues yet, since you can use other simpler matrix norms instead.)

[2] (Here are my original approaches to QR and LU in R, using solve instead of the special cases forwardsolve and backsolve.)

# OLD VERSIONS OF QR AND LU
# (before using Nicholas's suggested improvements)

# For a single b:

# QR, first attempt:
system.time({
qr_decomp = qr(A, tol = 1e-10)
xhat_qr = qr.coef(qr_decomp, b)
})
##    user  system elapsed
##   0.004   0.000   0.003
(err_qr = norm(x - xhat_qr))
## [1] 4.18e-08
(res_qr = norm(b - A %*% xhat_qr))
## [1] 6.11e-16

# QR, Nicholas' suggestion:
system.time({
qr_decomp = qr(A, tol = 1e-10)
R = qr.R(qr_decomp)
xhat_qr = backsolve(R, qr.qty(qr_decomp, b))
})
##    user  system elapsed
##   0.009   0.000   0.023
(err_qr = norm(x - xhat_qr))
## [1] 4.18e-08
(res_qr = norm(b - A %*% xhat_qr))
## [1] 6.11e-16

# LU, first attempt:
system.time({
lu_decomp = lu.decomposition(A)
xhat_lu = solve(lu_decomp$U, solve(lu_decomp$L, b))
})
##    user  system elapsed
##   0.002   0.000   0.003
(err_lu = norm(x - xhat_lu))
## [1] 3.99e-08
(res_lu = norm(b - A %*% xhat_lu))
## [1] 1.11e-16

# For many b's:

# QR, first attempt:
system.time({
qr_decomp = qr(A, tol = 1e-10)
xhats_qr = apply(bs, 2,
function(bi) qr.coef(qr_decomp, bi))
})
##    user  system elapsed
##   0.179   0.002   0.182

# QR, Nicholas' suggestion:
system.time({
qr_decomp = qr(A, tol = 1e-10)
qr_R = qr.R(qr_decomp)
xhats_qr = apply(bs, 2,
function(bi) backsolve(qr_R, qr.qty(qr_decomp, bi)))
})
##    user  system elapsed
##   0.157   0.000   0.159

# LU, first attempt:
system.time({
lu_decomp = lu.decomposition(A)
bs = apply(bs, 2,
function(bi) solve(lu_decomp$U, solve(lu_decomp$L, bi)))
})
##    user  system elapsed
##   0.167   0.001   0.173

### 8 responses to ““Don’t invert that matrix” – why and how”

1. baptiste

I don’t get why you don’t use solve(A, bs) with bs a matrix

• Good question. If you have all the b’s at once, then yep, that’s probably best.

I wanted to illustrate a situation where you have to reuse the same matrix A at different times with different b’s. (Imagine you’re solving this problem for new datasets that come in at different times.)

2. Nicholas

Hi Jerzy,
Good post. A few easy to implement suggestions…. You can get a little better with your qr time by saving on R like so:

system.time({
qr_decomp = qr(A, tol = 1e-10)
R = qr.R(qr_decomp)
xhats_qr = apply(bs, 2, function(bi) backsolve(R,qr.qty(qr_decomp, bi)))})

but more importantly, you can really speed up the lu calculation by using forwardsolve and backsolve:

system.time({
lu_decomp = lu.decomposition(A)
bs = apply(bs, 2, function(bi) backsolve(lu_decomp$U, forwardsolve(lu_decomp$L, bi)))
})

That becomes comparable in timing to repeated use of solve.

• Thanks! Great points. Somehow I didn’t realize R had forwardsolve and backsolve functions. (I’m surprised the qr documentation doesn’t link to them!)

I guess the generic solve must have tons of overhead that can be skipped if you already know the matrix structure and can use forwardsolve or backsolve directly.

I’ll update the post with your suggestions.

3. bk

Thanks for demonstrating why it’s perfectly OK to just invert the matrix and save our cognitive effort for bigger problems. You nicely demonstrate that even for a 7×7 matrix, which in the 1960s would have taken literally a day to invert, the difference between direct inversion and the smarter means of solving the linear equation saved 0.001 seconds in the best case. The errors are noticeably larger via inversion, but when err_inverting = 1.64e-07 on data of order of magnitude 1e-1, situations where we would care will not come frequently.

Finally, you add the comment that “If you really need to report these variances and covariances, I suppose you really will have to invert the matrix.” Given that the rule of thumb for when to report confidence intervals is “always”, you once again show why the inversion method is the norm and other solution methods reserved for special-case situations.

• Heh, good point. I just wanted to use a small simple example for illustration, but as you say, it doesn’t result in a difference that’s practically worth worrying about here.
I should link to a better real-world example, where the choice of inversion vs direct solving had a substantive impact.

On your 2nd point, I agree you should always report CIs—but still you might not need to invert the matrix.
* In OLS, if you only need CIs for certain predictions or linear combinations, like $Var(\hat{\beta}_1 + \hat{\beta}_2) = Var(\hat{\beta}_1) + Var(\hat{\beta}_2) + 2Cov(\hat{\beta}_1, \hat{\beta}_2)$, this is equivalent to a quadratic form with our inverted matrix in the middle. Aren’t there other matrix factorizations you could use to compute this quadratic form, without computing the inverse directly?
* If you’re only reporting separate CIs for each $\hat{\beta}$ on its own, you only need the main diagonal of that matrix inverse. Surely there’s some clever way (using standard basis vectors?) to get just that diagonal without inverting the whole matrix.
* But if your $\hat{\beta}$ vector has many entries and you literally need to report all the variances and covariances themselves, then yes, you do have to invert.

4. Nicholas

You can get variances and covariances from the R factor of the qr (or cholesky), again using forwardsolve-backsolve. Never a need for solve. I don’t know if it’s possible to get diagonals without the whole covariance matrix, though. It sure would be nice if you could! I find that all of the considerations you mention become especially important in large sparse applications (such as MCMC with Markov Random Fields), because the sparse Cholesky is a marvelous thing. I don’t know of a stats-friendly, applied, tutorial and introduction to these considerations in the sparse context, especially with permutation matrices, which always seem to trip one up.

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